A person on a 50-meter tall building throws a ball straight up into the air. This ball has an initial velocity of 20 m/s. From this:
1) Find the time it would take to reach the Max Height
2) Find the actual Max Height (In Meters)
3) Find the time it takes to return to the person
4) Find the velocity before it reaches the bottom ground.
at the max height, vy=0
vy=vi-gt set vy to zero, solve for t.
Height? Use that time
hf=hi+vit-1/2 gt^2
use the same equation, hf=hi, solve for t.
hf=-50 and hf=vi*t -1/2 gt^2 solve for time
To find the answers to the given questions, we can use the equations of motion. In this case, we assume that there is no air resistance.
1) Find the time it would take to reach the max height:
To find the time it takes to reach the max height, we need to consider the initial velocity and acceleration due to gravity. The ball will reach its maximum height when its vertical velocity becomes zero.
Using the equation: vf = vi + at,
where vf is the final velocity, vi is the initial velocity, a is the acceleration, and t is the time, we can rearrange the equation to solve for time:
0 = vi + at
Since the ball is thrown straight up, the initial velocity (vi) is positive (+20 m/s) and the acceleration due to gravity (a) is negative (-9.8 m/s²) since it acts in the opposite direction.
Plug in the values:
0 = 20 - 9.8t
Rearranging the equation to solve for time (t):
9.8t = 20
t = 20 / 9.8
Simplifying, the time it would take to reach the max height is approximately 2.04 seconds.
2) Find the actual max height:
To find the actual max height, we can use the equation:
vf² = vi² + 2as,
where vf is the final velocity, vi is the initial velocity, a is the acceleration, and s is the displacement.
We want to find the displacement (s) at the max height when the vertical velocity becomes zero.
The final velocity (vf) is 0 m/s since the ball is momentarily at rest at the max height. The initial velocity (vi) is +20 m/s, and the acceleration (a) is -9.8 m/s² due to gravity.
Plug in the values:
0² = (20)² + 2(-9.8)s
0 = 400 - 19.6s
Simplifying, the displacement (s) at the max height is approximately 20.41 meters.
3) Find the time it takes to return to the person:
To find the time it takes for the ball to return to the person, we can use the same equation:
vf = vi + at
We want to find the time (t) when the ball reaches the ground, so the final velocity (vf) is the negative of the initial velocity (vi) since it's moving in the opposite direction.
The initial velocity (vi) is +20 m/s, the acceleration (a) is -9.8 m/s² due to gravity, and the final velocity (vf) is -20 m/s.
Plug in the values:
-20 = 20 - 9.8t
Rearranging the equation to solve for time (t):
9.8t = 40
t = 40 / 9.8
Simplifying, the time it takes to return to the person is approximately 4.08 seconds.
4) Find the velocity before it reaches the bottom ground:
To find the velocity before it reaches the ground, we can use the same equation:
vf = vi + at
We know that the initial velocity (vi) is +20 m/s and the acceleration (a) is -9.8 m/s² due to gravity. We want to find the final velocity (vf) when the displacement (s) is negative (going down towards the ground).
Plug in the values:
vf = 20 - 9.8t
Substitute the time (t) it takes to reach the max height (2.04 seconds) into the equation:
vf = 20 - 9.8(2.04)
Calculating, the velocity before it reaches the bottom ground is approximately -1.96 m/s, indicating that the ball is moving downward.