Question

(a) Calculate the height of a cliff if it takes 2.45 s for a rock to hit the ground when thrown straight up from the cliff with an initial velocity of 8.75 m/s.
(b) How long would it take to reach the ground if it is thrown straight down with the same speed?

Answer 1

An object in free fall motion is under the influence of gravitational force only

The height and time are;

(a) The height of the cliff, h ≈ 8.00 meters

(b) The time it takes the rock to rich the ground with the same speed going  downward is approximately 0.67 seconds

The reason the above values are correct are as follows:

The given parameters are;

The time it takes for the rock to hit the ground when thrown straight up from the cliff, t = 2.45 s

The initial velocity with which the rock is thrown, u = 8.75 m/s

(a) To find the height of the cliff

Solution

The kinematic equation of motion is s = y₀ + u·t + (1/2)·g·t²

The time it takes to maximum height, $t_{max}$ = 8.75/9.81

The time it takes to get back to the cliff edge when thrown = 2 × 8.75/9.81

The time taken to fall at the velocity of u downwards from the cliff edge, $t_{cliff}$, is given as follows;

$t_{cliff}$ = 2.45 - 2 × 8.75/9.81 ≈ 0.666

Height of cliff, h = 8.75×(2.45 - 2 × 8.75/9.81) + (1/2) × 9.81 × (2.45 - 2 × 8.75/9.81)² ≈ 8

The height of the cliff, h ≈ 8.00 meters

(b) The time it takes the rock to rich the ground when thrown straight down art the same speed, (calculates above) = $t_{cliff}$ ≈ 0.67 seconds

Learn more about free fall motion here:

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