A child throws his basketball upward from the ground with an initial velocity of 30 ft/sec.
What will be the speed of the ball when it reaches the ground? (I need help uwu)
well, clearly
height = -16t^2 + 30t + c, where c is the initial height of the ball when it is released. ( It would be rather difficult for a child to throw the ball upwards from ground level at appr 20 mph, but let's assume c = 0)
then
0 = -16t^2 + 30t
16t^2 - 30t = 0
t(16t-30) = 0
t = 0 (the moment she throws the ball)
or
t = 30/16 = 1.875 seconds
velocity = d(height)/dt = -32t + 30
when t = 1.875, velocity = -32(1.875) + 30 = -30 ft/s
Adjust for any value of c that was given.
where did 16 came from?
To find the speed of the ball when it reaches the ground, we can use the principle of conservation of energy. The initial kinetic energy of the ball will be equal to the potential energy when it reaches the ground.
We know that the initial velocity of the ball is 30 ft/sec, and let's assume that the time it takes for the ball to reach the ground is t seconds. We can use the equation of motion:
h = ut + (1/2)gt^2
where h is the height, u is the initial velocity, g is the acceleration due to gravity (32 ft/sec^2).
Since the ball starts from the ground (h = 0) and the ball reaches the ground, we can say:
0 = 30t - (1/2)gt^2
Rearranging the equation, we get:
(1/2)gt^2 = 30t
Dividing both sides by t:
(1/2)gt = 30
Simplifying the equation, we find:
gt = 60
Since we know that g is 32 ft/sec^2, we can substitute it back to find t:
32t = 60
t = 60/32
t ≈ 1.875 seconds
Now that we have the time it takes for the ball to reach the ground, we can find its speed using the formula:
v = u + gt
where v is the final velocity and u is the initial velocity.
v = 30 + (32 * 1.875)
v ≈ 30 + 60
v ≈ 90 ft/sec
Therefore, the speed of the ball when it reaches the ground will be approximately 90 ft/sec.