A ball is thrown straight downward with a speed of 0.50 meter per second from a height of 4.0 meters. What is the speed of the ball 0.70 second after it is released?
v(t)=v(o)+gt g=-9.8m/s^2, v(o)=-.5m/s
I frankly would also verify that the ball had not hit the ground in this time, sometimes these question are tricks.
That was good advice from BobPursley.
After 0.7s, the ball will have travelled 2.75 meters down, so it will not have hit the ground yet.
If it had hit the ground, you would need a "turnaround time" (probably very short) and "coefficient of restitution" to get the velocity.
To solve this problem, we can use the equations of motion to determine the speed of the ball 0.70 seconds after it is released.
The first equation we can use is the equation for the final velocity:
v = u + at
Where:
v = final velocity
u = initial velocity
a = acceleration
t = time
Given:
u = 0.50 m/s (initial velocity)
a = 9.8 m/s^2 (acceleration due to gravity)
t = 0.70 s (time)
We can substitute these values into the equation:
v = 0.50 m/s + (9.8 m/s^2)(0.70 s)
Calculating:
v = 0.50 m/s + 6.86 m/s
v ≈ 7.36 m/s
Therefore, the speed of the ball 0.70 seconds after it is released is approximately 7.36 meters per second.
To find the speed of the ball after 0.70 seconds, we can use the kinematic equation for the final velocity of an object in free fall:
v = u + gt
Where:
v is the final velocity,
u is the initial velocity,
g is the acceleration due to gravity (approximately 9.8 m/s^2),
t is the time.
In this case, the ball is thrown straight downward, so the initial velocity is -0.50 m/s (negative because it's downward) and the time is 0.70 seconds.
Plug in these values into the equation:
v = -0.50 + 9.8 * 0.70
v = -0.50 + 6.86
v = 6.36 m/s
Therefore, the speed of the ball 0.70 seconds after it is released is 6.36 meters per second.