Question #226730
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.
1
Expert's answer
2021-08-16T14:44:13-0400

Given:

v0x=v0cosθv_{0x}=v_0\cos\theta

v0y=v0sinθv_{0y}=v_0\sin\theta

v0=30m/sv_0=30\:\rm m/s

y0=2.0my_0=2.0\:\rm m

θ=45\theta=45^{\circ}


The equations of motion of a car

x(t)=v0xt=30cos45t=21tx(t)=v_{0x}*t=30\cos45^{\circ}*t =21ty(t)=y0+v0ytgt22=2.0+30sin459.8t22=2.0+21t4.9t2y(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+21t4.9t2=02.0+21t-4.9t^2=0

t=4.4st=4.4\:\rm s

(b)

L=x(4.4s)=214.4=93mL=x(4.4\:\rm s)=21*4.4=93\:\rm m


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