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

Question

A 0.61 kg, 0.27 m diameter ball is thrown upward at 31 m/s 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?



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?


s


m


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?


s


m

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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