A ball is thrown straight up at a launch velocity of 15m/s. What is the velocity when it returns to the throwers hand?
you could do the equations OR
you could just recognize that (conservation of energy) the ball will slow to a stop and fall from that point and then achieve the same velocity it had at the original point
so, 15 m/s
To find the velocity of the ball when it returns to the thrower's hand, we need to understand the physics behind projectile motion.
When the ball is thrown straight up, it experiences a constant acceleration due to gravity in the downward direction. This acceleration is commonly represented as -9.8 m/s² since it acts opposite to the positive direction.
The velocity of the ball can be calculated using the equation of motion:
v = u + at
Where:
v is the final velocity
u is the initial velocity
a is the acceleration
t is the time taken
In this case, the initial velocity (u) is 15 m/s (given) and the acceleration (a) is -9.8 m/s² (due to gravity acting against the ball's motion).
When the ball reaches its maximum height, its velocity becomes zero momentarily. At this point, it starts falling back down due to the gravitational acceleration.
The time it takes for the ball to reach its maximum height and return to the thrower's hand is the same. This means that the total time in the air is twice the time taken to reach the maximum height.
To find the time of flight (t), we can use the equation:
v = u + at
Since the final velocity (v) is zero at the maximum height, we can rearrange the equation:
t = (v - u) / a
Substituting the known values, we get:
t = (0 - 15) / (-9.8)
t = 1.53 seconds (approximately)
Now that we have the time of flight, we can find the velocity when the ball returns to the thrower's hand.
v = u + at
v = 15 + (-9.8) * 1.53
v ≈ -14.95 m/s
Here, the negative sign indicates that the velocity is in the opposite direction of the initial velocity. So, the velocity of the ball when it returns to the thrower's hand is approximately -14.95 m/s.