A ball is thrown upward and returns to thrower hand in 12 seconds. What is speed with which ball is thrown and maximum height attained.
initial speed:
vf=vi+g5
but vf=-vi
-vi=vi-9.8t^2
vi=4.9*12^2
height:
at the top, vf=0
vf^2=vi^2+2ad
d=2vi/9.8 d is the same as hf
To find the speed with which the ball is thrown, we consider the fact that the vertical component of velocity is constant throughout the motion, disregarding air resistance.
Using this information, we can divide the total time into two equal parts: the time taken to reach the maximum height and the time taken to descend from the maximum height back to the thrower's hand.
Since the ball reaches its maximum height halfway through the total time, we can determine that the time taken to reach the maximum height is 12 seconds divided by 2, which equals 6 seconds.
First, let's find the speed with which the ball is thrown:
Acceleration due to gravity, g = 9.8 m/s² (assuming no air resistance)
When the ball reaches its highest point, its velocity becomes zero. From this point, it takes the same amount of time, 6 seconds, to descend back to the thrower's hand.
Using the equation vf = vo + gt, where vf is the final velocity, vo is the initial velocity, g is the acceleration due to gravity, and t is the time:
0 = vo + (-9.8) * 6
Simplifying the equation, we have:
0 = vo - 58.8
vo = 58.8 m/s
Hence, the initial speed with which the ball is thrown is 58.8 m/s.
Next, let's find the maximum height attained by the ball. We know that when the ball reaches its highest point, its vertical velocity becomes zero.
Using the equation vf = vo + gt, where vf and vo are the final and initial velocities, g is the acceleration due to gravity, and t is the time:
0 = vf + (-9.8) * 6
Simplifying the equation, we have:
0 = vf - 58.8
vf = 58.8 m/s
To find the maximum height, we use the equation:
vf² = vo² + 2as
Where vf is the final velocity (0 m/s), vo is the initial velocity (58.8 m/s), a is the acceleration (-9.8 m/s²), and s is the displacement (maximum height, h).
0 = (58.8)² + 2 * (-9.8) * s
Simplifying the equation, we have:
0 = 3456 + (-19.6) * s
-3456 = -19.6 * s
s = 176.33 m
Therefore, the maximum height attained by the ball is approximately 176.33 meters.