1. A motorcycle stunt driver rides his motorcycle off a small ledge that is 8.4 meters high. His initial velocity as he rides off the cliff is

24 m/s in the horizontal direction.

a. How long does it take before he lands on the ground below?

b. How far away from the base of the cliff does he travel while in

the air?

a. The time it took before the motorcycle stunt driver lands on the ground below is 1.31 seconds.

b. The distance from the base of the cliff that the motorcycle stunt driver traveled while in  the air is 31.44 meters.

Given the following data:

We know that the acceleration due to gravity (g) of an object on planet Earth is equal to 9.8

a. To determine how long (time) it took before the motorcycle stunt driver lands on the ground below:

At maximum height, time is given by the formula:

Where:

Substituting the given parameters into the formula, we have;

Time = 1.31 seconds.

b. To determine how far (distance) away from the base of the cliff the motorcycle stunt driver traveled while in  the air:

Horizontal distance = 31.44 meters.

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Explanation:

Given

Height S = 8.4m

Initial velocity u = 24m/s

Required

Time it will take to reach the ground

Using the equation of motion

S = ut +1/2gt²

8.4 = 24t + 1/2(9.8)t²

8.4 = 24t+4.9t²

4.9t²+24t - 8.4 = 0

Multiply through by 10

49t²+240t-84 = 0

Factorize

t = -240±√240²-4(49(-84)/2(49)

t = -240±√57600+16464/98

t = -240±√74064/98

t = -240±272.15/98

t = 32.15/98

t = 0.328seconds

Hence it takes 0.328seconds before it lands on the ground below.

b) The horizontal distance is the range expressed as;

R = u²/g

R = 27²/9.8

R = 729/9.8.

R = 74.39m

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