Answer: 16.3 seconds
Explanation: Given that the
Initial velocity U = 80 ft/s
Let's first calculate the maximum height reached by using third equation of motion.
V^2 = U^2 - 2gH
Where V = final velocity and H = maximum height.
Since the toy is moving against the gravity, g will be negative.
At maximum height, V = 0
0 = 80^2 - 2 × 9.81 × H
6400 = 19.62H
H = 6400/19.62
H = 326.2
Let's us second equation of motion to find time.
H = Ut - 1/2gt^2
Let assume that the ball is dropped from the maximum height. Then,
U = 0. The equation will be reduced to
H = 1/2gt^2
326.2 = 1/2 × 9.81 × t^2
326.2 = 4.905t^2
t^2 = 326.2/4.905
t = sqrt( 66.5 )
t = 8.15 seconds
The time it will take for the rocket to return to ground level will be 2t.
That is, 2 × 8.15 = 16.3 seconds
Hi there!

We know that:

U = Potential Energy (J)
K = Kinetic Energy (J)
E = Total Energy (J)
At 10m, the total amount of energy is equivalent to:
U + K = 50 + 50 = 100 J
To find the highest point the object can travel, K = 0 J and U is at a maximum of 100 J, so:
100J = mgh
We know at 10m U = 50J, so we can solve for mass. Let g = 10 m/s².
50J = 10(10)m
m = 1/2 kg
Now, solve for height given that E = 100 J:
100J = 1/2(10)h
100J = 5h
<u>h = 20 meters</u>