PRACTICE IT Use the worked example above to help you solve this problem. A ball

Question

PRACTICE IT 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 21.7 m/s straight upward, at an initial height of 50.9 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. 2.21 (b) Determine the maximum height. 23.876 (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. Time 4.43 Velocity -21.71 m/s (d) Determine the time needed for the ball to reach the ground. 6.12 (e) Determine the velocity and position of the ball at t-5.19 s Velocity Position m/s EXERCISE HINTS: GETTING STARTED II'M STUCK! A projectile is launched straight up at 65 m/s from a height of 83 m, at the edge of a sheer cliff. The projectile falls, just missing the cliff and hitting the ground below (a) Find the maximum height of the projectile above the point of firing (b) Find the time it takes to hit the ground at the base of the cliff. (c) Find its velocity at impact. m/s

Explanation / Answer

exercise:

initial speed=u=65 m/s

initial height=83 m

part a:

at maximum height, speed=0

using the formula,

final speed^2-initial speed^2=2*acceleration*distance

==>0^2-65^2=2*(-9.8)*distance

==>distance=215.56 m

part b:

let time taken be t seconds.

displacement to ground=-83 m

using the formula:

displacement=initial speed*time+0.5*acceleration*time^2

==>-83=65*t-0.5*9.8*t^2

==>4.9*t^2-65*t-83=0

solving for t, we get

t=14.4385 seconds

part c:

velocity at impact=initial speed+acceleration*time

=65-9.8*14.4385

=-76.497 m/s

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