A projectile is shot from the edge of a cliff 125 m above ground level with an initial speed of 65.0 m/s at an angle of 37 degrees with the horizontal

asked 2020-12-21
A projectile is shot from the edge of a cliff 125 m above ground level with an initial speed of 65.0 m/s at an angle of 37 degrees with the horizontal. (a) determine the time taken by the projectile to hitthe point P at ground level. (b) determine the range X of the projectile as measured from the base of the cliff at the instant just before the projectile hits point P. Find (x) the horizontal and vertical components of its velocity and (d) the magnitude of the velocity and (e) the angle made by the velocity vector with the horizontal (f) Find the maximum height above the cliff top reached by the projectile.

Answers (1)

a)Find the time until projectile hits the ground
Use Y components
\(\displaystyle{V}_{{i}}={65.0}{\sin{{37}}}\) make sure your calculator is set to degrees
\(\displaystyle{g}=-{9.8}\ \frac{{m}}{{s}^{{2}}}\)
Solve for T
b)To find range use X coordinates
a=0 this is always the case in projectile motion
Plug T from a) and find \(\displaystyle{X}_{{f}}\)
c)Remember there is no air friction on this problem, so accelaration is zero. That means velocity stays the same on the X coordinate.
As for the Y coordinate \(\displaystyle{V}_{{f}}={V}_{{i}}+{a}{T}\)
You know T, a=-g, and \(\displaystyle{V}_{{i}}\)
The direction is down or course.
d)The magnitude is found by \(\displaystyle\sqrt{{{y}^{{2}}+{x}^{{2}}}}\) you know the drill
e) I think this is \(\displaystyle\theta={\arccos{}}\) (x/magnitude found in d)
f) \(\displaystyle{V}_{{f}}={V}_{{i}}+{a}{t}\) use Y components
\(\displaystyle{V}_{{f}}={0}\) at the top
we know \(\displaystyle{V}_{{i}}\) and a=-g
find this T and use in the position formula to get \(\displaystyle{Y}_{{f}}\)
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