Question

a ball is thrown from the top of a building with an initial velocity of 28.0 m/s straight upwards,at an initial height of 25.0 m above the ground.the ball just misses the edge of the roof on its way down.determine;
(1) the time needed for the ball to reach its maximum height.
(2) the maximum height
(3) the time needed for the ball to return to the height from which it was thrown and the velocity of the ball at the instant

Answer 1

(1) To find the time needed for the ball to reach its maximum height, we can use the kinematic equation:
v = u + at
where
u = initial velocity = 28.0 m/s (upwards)
a = acceleration due to gravity = -9.81 m/s^2 (since the ball is moving upwards)
v = final velocity at max height = 0 m/s

Rearranging the equation to solve for time (t):
0 = 28.0 - 9.81t
t = 28.0 / 9.81 = 2.853 seconds

So, the ball reaches its maximum height after 2.853 seconds.

(2) To find the maximum height, we can use the kinematic equation:
s = ut + 0.5at^2
where
s = maximum height
u = initial velocity = 28.0 m/s (upwards)
a = acceleration due to gravity = -9.81 m/s^2
t = time to reach max height = 2.853 seconds

Plugging in the values:
s = 28.0(2.853) + 0.5(-9.81)(2.853)^2
s = 40.0 meters

Thus, the maximum height the ball reaches is 40.0 meters.

(3) To find 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, we can use the equation:
v = u + at

First, find the time it takes for the ball to return to the initial height:
s = ut + 0.5at^2
25 = 28t + 0.5(-9.81)t^2
0.5(-9.81)t^2 + 28t - 25 = 0

Solving this quadratic equation, we get t = 2.441 seconds

Now, to find the velocity of the ball at that instant, use the equation:
v = u + at
v = 28 - 9.81(2.441)
v = 3.55 m/s

Therefore, it takes 2.441 seconds for the ball to return to the initial height and its velocity at that instant is 3.55 m/s.