A ball is thrown from the top of a building with an initial velocity of 28.0m/s straight upwards, at an initial height of 25.0m 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 it's 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 that instant
1. To determine the time needed for the ball to reach its maximum height, we can use the kinematic equation:
v_f = v_i + at
where:
v_f = final velocity (0 m/s at maximum height)
v_i = initial velocity (28.0 m/s)
a = acceleration due to gravity (-9.81 m/s^2)
t = time
At the maximum height, the final velocity is 0 m/s, so the equation becomes:
0 = 28.0 - 9.81t
Solving for t:
t = 28.0 / 9.81
t ≈ 2.85 seconds
2. To determine the maximum height reached by the ball, we can use the kinematic equation:
s = v_i*t + (1/2)a*t^2
where:
s = maximum height
v_i = initial velocity (28.0 m/s)
t = time to reach maximum height (2.85 seconds)
a = acceleration due to gravity (-9.81 m/s^2)
Plugging in the values:
s = 28.0*2.85 + (1/2)*(-9.81)*(2.85)^2
s ≈ 40.13 meters
Therefore, the maximum height reached by the ball is approximately 40.13 meters.
3. The time needed for the ball to return to the height from which it was thrown can be determined by using the equation for total time of flight for a projectile:
total time = 2*t
total time = 2*2.85
total time = 5.7 seconds
To determine the velocity of the ball at that instant, we can use the kinematic equation:
v_f = v_i + at
where:
v_f = final velocity
v_i = initial velocity (28.0 m/s)
a = acceleration due to gravity (-9.81 m/s^2)
t = total time of flight (5.7 seconds)
Plugging in the values:
v_f = 28.0 - 9.81*5.7
v_f ≈ -33.48 m/s
Therefore, the time needed for the ball to return to the height from which it was thrown is 5.7 seconds, and the velocity of the ball at that instant is approximately -33.48 m/s. The negative sign indicates that the ball is moving downwards.