Question

a ball is thrown with a speed of 24.5m/s.after what time it will be at a height of 29.4m from the ground.?

Answer 1

Answer:

the ball is subjected to gravitational pull due to earth, so deceleration due to gravity is 9.8 m/sq. second.

so first you must calculate the time at which the velocity will become zero, because it is the highest point.let u be the initial velocity.

if t is the time then t=u/g=(24.4 m/s)/g=2.5 sec(approx.)

now calculate the height i.e at what point ball’s velocity becomes zero,

that height will be

H=(u*t)-(0.5*g*t*t)=(24.4*2.5)-(0.5*9.8*2.5*2.5)=61–3.125=57.875 m(approx.)

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