Question

While visiting the planet Mars, Moe leaps straight up off the surface of the Martian ground with an initial velocity of 105 m/s to catch a sleeping juicy Martian cricket. It took Moe 56 seconds to reach the bug. The gravitational acceleration on Mars is 3.75 2 m / s . Find how far the cricket was off the ground when it became Moe’s lunch. Must use integration to derive your results.

Answer 1

Answer:

L = 0 m

Therefore, the cricket was 0m off the ground when it became Moe’s lunch.

Explanation:

Let L represent Moe's height during the leap.

Moe's velocity v at any point in time during the leap is;

v = dL/dt = u - gt .......1

Where;

u = it's initial speed

g = acceleration due to gravity on Mars

t = time

The determine how far the cricket was off the ground when it became Moe’s lunch.

We need to integrate equation 1 with respect to t

L = ∫dL/dt = ∫( u - gt)

L = ut - 0.5gt^2 + L₀

Where;

L₀ = Moe's initial height = 0

u = 105m/s

t = 56 s

g = 3.75 m/s^2

Substituting the values, we have;

L = (105×56) -(0.5×3.75×56^2) + 0

L = 0 m

Therefore, the cricket was 0m off the ground when it became Moe’s lunch.