A 0.61 kg, 0.27 m diameter ball is thrown upward at 31 from the origin at yo = 0

Question

A 0.61 kg, 0.27 m diameter ball is thrown upward at 31 from the origin at yo = 0.


a) At what time did the ball reach its highest position?


s


b) What was the ball's highest position?


m


c) What were the ball's time and velocity when the ball was on the way up and the position was half the distance calculated in part (b) above?



In sec



In m/s


d) What were the ball's time and velocity when the ball was on the way down and the position was half the distance calculated in part (b) above?



In s



m/s


e) What were the time and position when the ball was on the way up and had a velocity equal to half of the initial velocity?


In sec


In meters


f) What were the time and position when the ball was on the way down and had velocity equal to half of the initial velocity?

In Sec


In meters

Explanation / Answer

solve: mass of the ball m=0.61 kg diameter of the ball =0.27 m the initial velocity of the ball v_0 =31 m/s the final velocity of the ball v=0 m/s initial position of the ball y_0 =0 a) we know, t=v-v_0/a time t =(0-31 m/s)/(-g) t =(31m/s)/(9.8 m/s^2) t =3.16 s b) maximum height y =v^2-v_0^2/2a y =(0 m/s)^2-(31 m/s)^2/2(-9.8 m/s^2) =49.03 m c) time t=v-v_0/a t=v_0/g t =(31 m/s)/(9.8 m/s^2) =3.16 s we know , v^2-v_0^2 =-2g(y) where y =49.03 m/2 v^2-v_0^2 =-2g(y/2) velocity v =21.92 m/s d) y/2=v_0t+1/2 gt^2 49.03 m/2 =1/2(9.8 m/s^2)t^2 time t =2.23 s we know ,v^2-v_0^2 =-2g(y/2) velocity v =21.92 m/s e) time t =v-v_0/a t=[(v_0/2) -v_0]/(-g) t =v_0/2g =31 m/s/2(9.8 m/s^2) =1.58 s position y = 3v_0^2/8 g =12.25 m f)time t =v-v_0/a t=[(v_0/2) -v_0]/(g) t =v_0/2g =31 m/s/2(9.8 m/s^2) =1.58 s velocity v =v_0/2 v_0 =2v position y = -(v_0/2)^2/2g =-12.25 m

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