Question

A distant planet has an acceleration due to gravity of 4 m/s² near its surface. An object is released from rest from the top of a tall cliff on the planet, and the object lands at the bottom of the cliff in 20 seconds. A second object is then thrown upward from the edge of the same cliff with a speed of 4 m/s. The time it takes the second object to reach the bottom of the cliff is most nearly

a. 19 s
b. 21 s
c. 22 s
d. 80 s

Answer 1

The second object will take 1 second to cease its upward motion, after which it will fall for the same 20 seconds as the first object did from rest. The total time for the second object to hit the bottom of the cliff is therefore 21 seconds.

Step-by-step explanation:

To solve for the time taken by the second object to reach the bottom of the cliff, we first determine how long it will take the object to stop its upward motion under the planet's gravity. We use the formula v = u + at, where v is the final velocity (0 m/s at the peak of its travel), u is the initial velocity (4 m/s), a is the acceleration due to gravity (-4 m/s²), and t is the time. Solving for t, we get:

t = (v - u) / a = (0 - 4) / -4 = 1 s

It takes 1 second to stop the upward motion. Then, the object will start falling from rest. We already know from the first object that it takes 20 seconds to fall to the bottom from rest. Therefore, the total time required for the second object to reach the bottom will be the time to stop plus the fall time:

Total time = 1 s (upward motion time) + 20 s (fall time from rest) = 21 s

So the correct answer is (b) 21 s.

Answer 2

Using kinematic equations for uniformly accelerated motion, the time it takes the second object, thrown upward with an initial speed of 4 m/s, to reach the bottom of an 800-meter cliff on a planet with gravity 4 m/s² is approximately 21 seconds. Therefore, the correct answer is option b. 21 s.

Step-by-step explanation:

To find the time it takes for the second object to reach the bottom of the cliff after being thrown upward, we'll use the kinematic equations for uniformly accelerated motion:

v = u + at
s = ut + ½at²
v² = u² + 2as

Where:

• v is the final velocity,
• u is the initial velocity,
• a is the acceleration due to gravity,
• t is the time,
• s is the displacement.

For the first object, which falls from rest:

0 + 0.5 × 4 × (20)² = 800 m

This means the cliff is 800 meters tall. For the second object that is thrown upwards:

It will first rise to a certain height where its velocity becomes 0 m/s. Using the first equation, we can find the time t to reach this height:

0 = 4 + (-4) × t
t = 1 second

During this 1 second of upward motion, the object rises:

4 × 1 + 0.5 × (-4) × (1)² = 2 m

The remaining height to fall is therefore 800 m - 2 m = 798 m. The total time for the upward motion and free fall is the solution to:

798 = 0 × t + 0.5 × 4 × t²
t² = 399
t = √399 ≈ 20 seconds

Thus, the total time is roughly 1 second (upward) + 20 seconds (falling) = 21 seconds. Therefore, the correct option is b. 21 s.