a tennis ball is hit vertically upward with a speed of 21.0m/sec from the roof of a restaurant that is 83.8 m above the ground. how long after being hit does the ball hit the ground?

Question

mos1635 · Answer

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equation for deaccelerated motion???

mos1635 · Answer

can you wright it??

anonymous · Answer

vy=v0-gt

mos1635 · Answer

that is for velocity
and tho one for dispacement?

mos1635 · Answer

displacement

anonymous · Answer

From the preservation of the kinetic energy theorem we get that F=W where F is the force applied to the body and W the weight of the body.so the body is decelerating with a=g.At some point the body has velocity v=0 from which we get that v0=a*t => v0=g*t => t=21/9.81=2.14s.then using the formula h=1/2*a*t^2 => h=1/2*g*t^2 we get h=22.46m.So the body is now in a height of 22.46+83.8=106.26m.Then from the free drop formula t=sqrt(2*h/g) equals to approximately 4.65s.From that we get that the total time before the ball hit the ground is T=t1+t2=4.65+2.14=6.79s and rounding to get two important digits T=6.8s.I think it is correct but any correction is accepted.

anonymous · Answer

why did you make the vo of h=vot+1/2at^2 zero

anonymous · Answer

v0=21m/s but the velocity of the body when it reaches the maximum height is 0.