A student drops a ball from the top of a tall building; the ball takes 2.4 to reach the ground. What was the ball’s speed just before hitting the ground? (this is excluding everything like air resistance, etc)
Assuming it takes 2.4 seconds to reach the ground, the final speed is
2.4*9.81 m/sec²
=23.5 m/s.
In fps units, multiply by 32.2 ft/sec².
To find the speed of the ball just before it hits the ground, we can use the equation:
v = gt
where:
v = velocity or speed
g = acceleration due to gravity (approximately 9.8 m/s^2)
t = time taken to reach the ground
In this case, the time (t) is given as 2.4 seconds.
Let's calculate the speed:
v = (9.8 m/s^2) * (2.4 s)
v ≈ 23.52 m/s
Therefore, the ball's speed just before hitting the ground (excluding air resistance) is approximately 23.52 m/s.
To find the speed of the ball just before hitting the ground, you can use the equation of motion for an object in free fall, considering the acceleration due to gravity (-9.8 m/s^2) and the time it takes to reach the ground (2.4 seconds). The equation is:
v = u + at
Where:
v is the final velocity (speed) of the ball just before hitting the ground,
u is the initial velocity (speed) of the ball,
a is the acceleration due to gravity (approximately -9.8 m/s^2), and
t is the time taken (2.4 seconds).
Since the ball is dropped (not thrown), the initial velocity is 0 since it starts from rest. Therefore, the equation becomes:
v = 0 + (-9.8 m/s^2) x (2.4 s)
Simplifying, we have:
v = -23.52 m/s
However, since speed is a scalar quantity (magnitude without direction), we take the absolute value of the velocity to get the speed. Therefore, the speed of the ball just before hitting the ground is approximately 23.52 m/s.