A car slid off an icy 10m bridge and landed 12m away from the bridge. How much time was the car in the air? (Hunt: Projectile)

Respuesta :

You just need to use the height of the bridge from the ground.

The formula to be used is y = Voy*t + g *(t^2) / 2, where the Voy is the intitial vertical velocity, which is zero.

Then, y = 10 m = g * (t^2) / 2 => t^2 = 2 * 10 m / g

=> t^2 = 2 * 10 m / 9.8 m/s^2 = 2.04 s^2

=> t = 1.43 s <------- answer


You will use the height of the bridge from the ground.

Solution:

Formula to be used is y=Viy(t)+g(t^2)/2

Where:

Vi=initial velocity which is 0 m/s

 y=10 m

Gravitational acceleration or g =9.8m/s^2

T= time you need

Substitute all the given to the formula

10m=(0m/s)(t)+(9.8m/s^2)(t^2)/2

10mx2=9.8m/s^2(t^2)

Now isolate the variable you want to find which is T or time

10mx2/9.8m/s^2=t^2

20m/9.8m/s^2=t^2

Square root of 2.04= square root of t^2

T=1.43 secs

The answer is 1.43 seconds