part 1 A ball is thrown upward. After reaching a maximum height, it continues fa

Question

part 1

A ball is thrown upward. After reaching a maximum height, it continues falling back to- ward Earth. On the way down, the ball is caught at the same height at which it was thrown upward.
If the time (up and down) the ball re- mains in the air is 2.18 s, find its speed when it caught. The acceleration of grav- ity is 9.8 m/s2 . Neglect air resistance.
Answer in units of m/s

part 2
If the time the ball remains in the air is 2.31 s, find the maximum height hmax the ball at- tained while in the air.
Answer in units of m

Explanation / Answer

Part 1 if the ball is coming downward a=g=9.8 m/s2 the time for going upwards = the time for going downwards so the time for going downwards t = 2.18/2 = 1.09 sec Vi = 0 here we use Vf = Vi + gt Vf = 0 + (9.8)(1.09) Vf = 10.68 m/s Part 2 the time for going upwards = the time for going downwards so the time for going downwards t = 2.31/2 = 1.155 sec for the journey of moving downwars a=-9.8 m/s2 t= 1.155 sec u = 0 the height fallen = ut + 0.5 * a *t2 = 0.5 * -9.8 * 1.155^2 =-6.54 m so hmax is same as height fallen hence hmax = 6.54 m

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