PLEASE ANSWER FULL QUESTION CORRECT FOR POINTS*** A stone is thrown from the top

Question

PLEASE ANSWER FULL QUESTION CORRECT FOR POINTS*** A stone is thrown from the top of a building with an initial velocity of 20.5 m/s straight upward, at an initial height of 52.5 m above the ground. The stone just misses the edge of the roof on its way down, as shown in the figure.

(a) Determine the time needed for the stone to reach its maximum height.
s

(b) Determine the maximum height.
m

(c) Determine the time needed for the stone to return to the height from which it was thrown, and the velocity of the stone at that instant.

(d) Determine the time needed for the stone to reach the ground.
s

(e) Determine the velocity and position of the stone at t = 6.88 s.

m

Explanation / Answer

v=u+at v=final velocity u=initial velocity a=accel t=time
v=0 u=20.5 a=-g=-9.8
t=u/g=20.5/9.8=2.09sec
s=height above roof edge
v2-u2=2as s=(v2-u2)/2a=-420.25/-19.6=21.44m
Max height=52.5+s
=52.5+21.44=73.94m

time needed for the stone to return to the height from which it was thrown=2*t
=2*2.09=4.18sec
because motion is symmetric

||ly
velocity at the instant=-initial velocity=-20.5m/sec
just the direction is reversed

Tground=4.18+sqrt(2Hcurrent/g)=4.18+sqrt(2*52.5/9.8)=7.45 s

v=-20.5-at=-20.5-9.8*(6.88-4.18)=-46.96 m/s

s=52.5-1/2*9.8*(6.88-4.18)^2=52.5-35.721=16.779 m

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