A golfer hits a golf ball at an angle of 25 degrees to the ground. If the golf ball covers a horizontal distance of 301. 5 m, what is the ball's maximum height? (Hint: at the top of its flight, the ball's vertical velocity component will be zero.)

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A golfer hits a golf ball at an angle of 25 degrees to the ground. If the golf ball covers a horizontal distance of 301. 5 m, what is the ball&#039;s maximum height? (Hint: at the top of its flight, the ball&#039;s vertical velocity component will be zero.)

BleabyinfibiaG · Accepted Answer

First off, Velocity V is equal to V=Vξ+Vyj where Vx=Vcos⁡(∠) and Vy=Vsin⁡(∠) since the golf ball covers a distance of 301.5m, we canuse Vx⋅Time=Distance[x] Or V=Distance[x]Time⋅cos⁡(∠) Height is a function of time and using the following equations ddt BSK velocity = acceleration, Vy[t]=Vy[t=0]−&amp;gt; and ddt distance = velocity Dy[t=t]=(Vy[t=0]⋅t−&amp;gt;22) to find the maximum height, Vy[t]=0, or t=Vyg. Rearranging, we get Dy=Vy2g−Vy2g2 or Dy=Vy22g To find the velocity, we need to use the fact that the highest point is exactly halfway between the point where the ball is hit and the point where the ball lands,  So t=2⋅Vyg And using our first equation, Vx⋅Time=Distance[x] Vx⋅2Vyg=301.5 V2⋅cos⁡(∠)⋅sin⁡(∠)⋅2g=301.5 Once you find V, just plug it back into Dy=Vy22g This will be your answer

Jeffrey Jordon · Accepted Answer

the horizontal displacement can be expressed as:  △x=vxt  Height is expressed as△y=vyt−12gt2  The time it would hit the maximum height is expressed as:  0=vy−gt→t=vyg  The time that the projectile reaches the ground is twice the time it takes the projectile to reach its maximum height. This is expressed as  t=2vyg  to find the time of the projectile to reach the maximum height, we have the formula  r=vx2vyg  From trigonometry,  vx=cos⁡(θ), vy=sin⁡(θ)  r=2cos⁡(θ)sin⁡(θ)g  We know from our trigonometric identities that  sin⁡(2θ)=2sin⁡(θ)cos⁡(θ) , so it would become  r=v2gsin⁡(2θ)  ymax=v2yg−12g(vyg)2=v2yg−12v2yg=v2sin⁡2(θ)2g  velocity is R=301.5  θ  =2⋅5g=10  m/s2  V(0)=? Y(max)=?  R=R=v2⋅sin⁡(2θ)g  Substitute the answers  in the equation v ll,  that would be  V2=3935 m/s we also know that at the highest point V(y)=0 so we v can write down:  V(y)2−V(0)(y)2=−2g△y  we also know that  V(y)=V(0)⋅sin⁡θ  V(y)2−[V(0)sin⁡θ]2=−2g△y  y :35m

user_27qwe · Accepted Answer

Step 1: Identify the given information:The angle of the golf ball&#039;s trajectory, θ=25 degrees.The horizontal distance covered by the ball, d=301.5 m.Step 2: Find the horizontal and vertical components of the initial velocity:The initial velocity, v0, can be split into its horizontal component, v0x, and vertical component, v0y.v0x=v0·cos(θ)v0y=v0·sin(θ)Step 3: Find the time of flight:The time it takes for the ball to reach its maximum height is equal to the time it takes for the ball to return to the ground. At the top of its flight, the vertical velocity component is zero. We can use this information to find the time of flight, t.v0y−g·t=0t=v0ygStep 4: Find the maximum height:The maximum height, h, can be calculated using the formula:h=v0y·t−12·g·t2Substituting the value of t from Step 3:h=v0y·v0yg−12·g·(v0yg)2Simplifying the equation:h=v0y22gStep 5: Substitute the values and calculate the maximum height:Substituting the values of v0y, θ, and g into the equation in Step 4, we can calculate the maximum height, h.h=(v0·sin(θ))22gSubstituting v0=dcos(θ):h=(dcos(θ)·sin(θ))22gh=d2·sin2(θ)2g·cos2(θ)Substituting the given values of d, θ, and g=9.8m/s2:h=(301.5)2·sin2(25∘)2·9.8·cos2(25∘)h≈29.82m―Therefore, the golf ball&#039;s maximum height is approximately 29.82 meters.

alenahelenash · Accepted Answer

Given:Angle of launch (θ) = 25 degreesHorizontal distance covered (d) = 301.5 mTo find: Maximum height (H) of the golf ballUsing the kinematic equations, we can analyze the motion of the golf ball. At the top of its flight, the vertical velocity component (vy) will be zero.We can decompose the initial velocity (v0) of the golf ball into its horizontal component (v0x) and vertical component (v0y):v0x=v0cos(θ)v0y=v0sin(θ)Since the ball&#039;s vertical velocity component is zero at the top of its flight, we can use the equation for vertical displacement (Δy) to find the maximum height (H):Δy=v0y·t−12gt2where t is the time taken to reach the maximum height and g is the acceleration due to gravity.At the top of its flight, the ball&#039;s vertical displacement is equal to the maximum height:H=12gt2To find the time taken to reach the maximum height, we can use the equation for horizontal displacement (d):d=v0x·tSolving for t, we get:t=dv0xSubstituting this value of t into the equation for maximum height, we have:H=12g(dv0x)2Now, we can substitute the given values and calculate the maximum height:H=12·9.8·(301.5v0cos(θ))2

karton · Accepted Answer

To solve this problem, we can use the equations of projectile motion. Let&#039;s denote the initial velocity of the golf ball as v0 and the angle of projection as θ.The horizontal and vertical components of the initial velocity are given by:V0x=v0cos(θ) (horizontal component)V0y=v0sin(θ) (vertical component)At the top of the ball&#039;s flight, the vertical velocity component is zero (Vy=0). We can use this information to find the time it takes for the ball to reach its maximum height.The vertical displacement (Δy) can be calculated using the equation:Δy=V0y·t+12·g·t2Since the vertical velocity at the top of the flight is zero, we have:0=V0y−g·tSolving this equation for t, we find:t=V0ygSubstituting this value of t into the equation for Δy, we get:Δy=V0y2g−12·g·(V0yg)2Simplifying this equation, we have:Δy=V0y22gNow, we can substitute the expressions for V0y and g:Δy=(v0sin(θ))22·9.8The maximum height is equal to the vertical displacement, so the maximum height (h) is given by:h=(v0sin(θ))22·9.8Given that the horizontal distance covered by the ball is 301.5 m, we can use the horizontal component of the initial velocity to find the initial velocity (v0).The horizontal displacement (Δx) is given by:Δx=V0x·tSubstituting the expression for V0x and the value of t from above, we have:Δx=(v0cos(θ))·(V0yg)Simplifying this equation, we get:Δx=(v0cos(θ))·(v0sin(θ))gSubstituting the given value of Δx:301.5=(v0cos(25∘))·(v0sin(25∘))9.8Simplifying this equation, we can solve for v0:v02=301.5·9.8cos(25∘)·sin(25∘)Taking the square root of both sides, we get:v0=301.5·9.8cos(25∘)·sin(25∘)Finally, substituting this value of v0 into the equation for the maximum height, we can calculate the maximum height (h):h=(301.5·9.8cos(25∘)·sin(25∘)·sin(25∘))22·9.8Calculating this expression will give us the value of the ball&#039;s maximum height.

A golfer hits a golf ball at an angle of 25 degrees to the ground. If the golf ball covers a horizontal distance of 301. 5 m, what is the ball's maximum height? (Hint: at the top of its flight, the ball's vertical velocity component will be zero.)

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