Question

A stunt man drives a car at a speed of 20 m/s off a 28-m-high cliff. The road leading to the cliff is inclined upward at an angle of 20°. How far from the base of the cliff does the car land? What is the car's impact speed?

Answer 1

Answer:

car land from base at distance 39.66 m

impact speed = 45.17 m/s

Explanation:

given data

speed v = 20 m/s

height h  = 28 m

angle θ = 20°

to find out

How far from the base of cliff does the car land and car impact speed

solution

we consider here x is horizontal direction and

and y is vertical direction

so we find time first for climb 28 m high

use equation of motion for climb height

- h = v×sinθ × t - 0.5× a×t²  ............1

here h is height, t is time , a is acceleration and v is speed

put here all value

-28 = 20×sin20 × t - 0.5× 9.8 ×t²

t = 4.89 seconds

so

distance = ut = 20×cos20 × t = 20×cos20 × 4.86

distance = 39.66 m

so car land from base at distance 39.66 m

and

for impact speed we calculate vx and vy

we use equation of motion

v = u - at   ..................2

at vertical direction

vy = 20×sin20 - 9.8 ( 4.89) = -41.08 m/s  

at horizontal direction

vx = 20×cos20

vx = 18.79 m/s

so impact speed = $\sqrt{vx^{2} + vy^{2}}$    .................3

put here value

impact speed = $\sqrt{18.79^{2} + (-41.08)^{2}}$

impact speed = 45.17 m/s