Use the worked example above to help you solve this problem. A ball is thrown fr

Question

Use the worked example above to help you solve this problem. A ball is thrown from the top of a building with an initial velocity of 20.9 m/s straight upward, at an initial height of 54.7 m above the ground. The ball just misses the edge of the roof on its way down, as shown in the figure (a) Determine the time needed for the ball to reach its maximum height. (b) Determine the maximum height. (c)Determine the time needed for the ball to return to the height from which it was thrown, and the velocity of the ball at that instant. (d) Determine the time needed for the ball to reach the ground. (e) Determine the velocity and position of the ball at t = 5.44 s.

Explanation / Answer

a) maximum height

v-u=at ;whrer v=o,o-u=at

t=u/a(time taken to reach max height)

=20.7/9.81

=2.110

b) using equation of motion

   h=ut=1/2at^2

=1/2*9.81*2.316

=11.35998 m

c) t=2t(time needed for the ball return to the height from which it was thrown=2times required to max height

t =2*2.1100

=4.22 sec

v will be upward so v =20.7 m/sec

e) given t =5.44 sec

ie taken initial position as if t = 5.44-4.22

=1.22 sec

now s =ut +1/2 at^2
= 20.7*1.22+1/2*9.8*1.038*1.038
=30.5334756 m

then v =u+at

=22.7+9.8*1.038

=32.87 m/sec

d) -54.7 =-22.7t - 1/2 at ^2

t = 1.7492 sec

total time = 1.7492 +4.632
=6.3812 sec

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