Question

a motorcycle is trying to leap across the canyon by driving horizontally off a cliff 38 m/s. Ignoring air resistance, find the speed with which the cycle strikes the ground on the other side

Answer 1

Answer:

46.15m/s

Explanation:

We are given that

Initial speed,u=38 m/s

Height of cliff,h=70 m

We have to find the speed with which the cycle strikes the ground at a height 35 m on the other side.

h'=35 m

Using conservation law of energy

$\frac{1}{2}mu^2+mgh=\frac{1}{2}mv^2+mgh'$

$\frac{1}{2}u^2+gh=\frac{1}{2}v^2+gh'$

Where $g=9.8m/s^2$

Substitute the values

$\frac{1}{2}(38)^2+9.8\times 70=\frac{1}{2}v^2+9.8\times 35$

$\frac{1}{2}v^2=\frac{1}{2}(38)^2+9.8\times 70-9.8\times 35=1065$

$v=\sqrt{2\times 1065}$

$v=46.15m/s$