Question

A car traveling at 30 m/s drives off a cliff that is 50 meters high? How far away does it land?

Answer 1

To find the distance at which the car will land, we can use the equation for distance in free-fall motion and the time it takes for the car to hit the ground. The car will land approximately 95.7 meters away from the cliff.

Step-by-step explanation:

To find the distance at which the car will land, we can use the equation for distance in free-fall motion:

d = v_i * t + 0.5 * a * t^2

Since the car is traveling horizontally, the initial vertical velocity is 0 m/s. The acceleration due to gravity is -9.8 m/s^2. We can use the equation to solve for the time it takes for the car to hit the ground:

50 m = 0 * t + 0.5 * (-9.8 m/s^2) * t^2

Simplifying, we get:

t^2 = 10.2

t = 3.19 s

Finally, we can use the time to find the horizontal distance traveled:

d = v * t

d = 30 m/s * 3.19 s = 95.7 m

Therefore, the car will land approximately 95.7 meters away from the cliff.

Answer 2

The maximum range
`R_(max)= 132. 72` m

Step-by-step explanation:

Given,

The initial velocity of the car, u = 30 m/s

The height of the cliff, h = 50 m

Let the car drives off the cliff with a horizontal velocity of 30 m/s.

The formula for a projectile that is projected from a height h from the ground is given by the relation

`R_(max)= (u)/(g)\sqrt{u^(2) + 2gh }` m

Where,

g - acceleration due to gravity

Substituting the values in the above equation

`R_(max)= (30)/(9.8)\sqrt{30^(2) + 2X9.8X50 }`

= 132.72 m

Hence, the car lands at a distance,
`R_(max)= 132. 72` m