A man throws a ball straight up in the air with an initial speed of 24 meter per second.
How long does it take for the ball to return to him?
What is the maximum height of the ball above the point where it was thrown?
The ball spends Vo/g = 2.45 seconds going up and an equal amount of time going back down. 2 x 2.45 = ___ s
Maximum height = (average velocity) x (time going up)
= 12 m/s * 2.45 s
= ___ m
4.9 SECONDS
29.4 M
To find out how long it takes for the ball to return to the man and the maximum height of the ball, we can use the equations of motion and kinematics.
1. Time taken for the ball to return to the man:
The time it takes for the ball to return to the same height can be found using the equation for vertical motion:
h = v₀t - (1/2)gt²,
where h is the height, v₀ is the initial velocity, t is the time, and g is the acceleration due to gravity (approximately 9.8 m/s²).
In this case, since the ball is thrown straight up, the final height (h) is the same as the initial height, and the final velocity (v) is -v₀ (negative because it is moving in the opposite direction).
Therefore, the equation becomes:
0 = v₀t - (1/2)gt².
Rearranging the equation and solving for t:
(1/2)gt² = v₀t,
(1/2)gt = v₀,
t = 2v₀/g.
Substituting the given values:
t = 2 * 24 m/s / 9.8 m/s².
t ≈ 4.9 seconds.
So, it takes approximately 4.9 seconds for the ball to return to the man.
2. Maximum height of the ball:
The maximum height can be found using the equation for vertical motion:
v = v₀ - gt.
At the maximum height, the ball reaches its peak, so its final velocity (v) is 0 m/s.
Therefore, the equation becomes:
0 = v₀ - gt.
Solving for t:
t = v₀ / g.
Substituting the given values:
t = 24 m/s / 9.8 m/s².
t ≈ 2.4 seconds.
To find the maximum height (h), we can use the equation:
h = v₀t - (1/2)gt².
Substituting the values for t:
h = 24 m/s * 2.4 s - (1/2) * 9.8 m/s² * (2.4 s)².
h ≈ 28.1 meters.
Therefore, the maximum height of the ball above the point where it was thrown is approximately 28.1 meters.