Answer to Question #226730 in Physics for manisha

Question

In a movie scene, a car was driven at a speed of 30 m s-1

up a ramp inclined at 45o
to the ground. The car ran off the top of the ramp which is 2.0 m above the ground.
Assuming that air resistance is negligible, calculate
a) the time the car is in the air.
b) the horizontal distance-travelled by the car from the ramp to the point of
impact on the ground.

Answer 1

Given:

$v_{0x}=v_0\cos\theta$

$v_{0y}=v_0\sin\theta$

$v_0=30\:\rm m/s$

$y_0=2.0\:\rm m$

$\theta=45^{\circ}$

The equations of motion of a car

$$x(t)=v_{0x}*t=30\cos45^{\circ}*t =21t$$ $$y(t)=y_0+v_{0y}*t-\frac{gt^2}{2}=2.0+30\sin45^{\circ}-\frac{9.8*t^2}{2}\\ =2.0+21t-4.9t^2$$

(a)

$$2.0+21t-4.9t^2=0$$

$$t=4.4\:\rm s$$

(b)

$$L=x(4.4\:\rm s)=21*4.4=93\:\rm m$$