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
To develop this problem, it is necessary to apply the concepts related to the description of the movement through the kinematic trajectory equations, which include displacement, velocity and acceleration.
The trajectory equation from the motion kinematic equations is given by
Where,
a = acceleration
t = time
= Initial velocity
= initial position
In addition to this we know that speed, speed is the change of position in relation to time. So
x = Displacement
t = time
With the data we have we can find the time as well
With the equation of motion and considering that we have no initial position, that the initial velocity is also zero then and that the acceleration is gravity,
Therefore the vertical distance that the ball drops as it moves from the pitcher to the catcher is 1.46m.
I added individual steps for clarity. Note that g must be positive if the solution is to be real.
Let me know if you have any questions.
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