# The Stunt Driver: Jason's Cliff Jump Calculation

## Jason's Stunt: Calculating Time and Distance

**Jason** is a stunt driver who jumps off a cliff with a speed of 25 m/s. The cliff's height is 30 meters. We need to determine how long he remains in the air and how far from the bottom of the cliff he lands safely.

Final Answer:

**Jason** remains in the air for approximately 2.47 seconds and lands 61.75 meters away from the bottom of the cliff.

Explanation:

In this scenario, we can use equations of motion under gravity to solve the problem. First, we calculate the time **Jason** is in the air. The formula to calculate time is derived from the equation of motion: *distance = initial velocity x time + 0.5 x acceleration x (time^2)*.

Given that the acceleration due to gravity is -9.8 m/s^2 and the initial velocity is 25 m/s, we use the formula: *t = sqrt((2*distance) / acceleration)*. Substituting the values, we get *t = sqrt((2*-30)/-9.8) ≈ 2.47 seconds*.

Secondly, to calculate how far **Jason** lands from the bottom of the cliff, we consider his horizontal movement. Since there is no horizontal acceleration, we use the formula *distance = speed x time*.

Therefore, *d = 25m/s x 2.47s = 61.75 meters*. This means that **Jason** lands 61.75 meters away from the bottom of the cliff.

Would you like to try solving a similar physics problem related to motion under gravity?

Yes, I would be interested in attempting a physics problem involving motion under gravity. Please provide me with the details.