The exit velocity of a baseball (its velocity as it leaves the bat) is 128 feat per second, in the direction of 30 above horizontal and directly cowards the monster 30 foot tall center field wall. If the ball was hit 4 feet above ground level, find

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Step 1 Parametric equations related two or more variables with one other variable. The one variable is known as an independent variable. The two or more variables that relate to the independent variable are known as dependent variables. A vector can be broken down into its components along the coordinate axes, that is along the x and the y axes. Step 2 The velocity vector makes an angle of 30∘ with the horizontal. Thus the component along the x-axis is obtained vcos⁡θ and the component along the y-axis is obtained as vsin⁡θ The distance covered along the x-axis by the ball will be equal to the velocity in the x-direction multiplied by the time variable. There is no acceleration in the x-direction. Thus the distance along the x-axis is obtained as follows: x=(vcos⁡θ)t =(128cos⁡30∘)t =110.85t ft Step 3 The displacement of the ball along the y-axis can be obtained by using the expression s=ut+12at2. The acceleration along the y-axis is −32 fts2 The initial distance from the ground of the ball is 4 ft. The initial velocity is the velocity component in the y-direction. Thus the displacement along the y-axis is obtained as follows: y=4+ut+12at2 =4+(vsin⁡θ)t+12×(−32t2) =4+(128sin⁡30∘)t−16t2 =−16t2+64t+4 ft Thus the parametric equations are obtained as follows: x=110.85tft y=−16t2+64t+4 ft

The exit velocity of a baseball (its velocity as it

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