Question

A skier leaves the horizontal end of a ramp with a velocity of 31.0 m/s and lands 156.3 m from the base of a ramp how high is the end of the ramp from the ground ?

Answer 1

Answer:

The height of ramp = 124.694 m

Explanation:

Using second equation of motion,

$s = ut + \frac{1}{2}at^2$

From the question,

u = 31 m/s; s = 156.3 m, a=0

substituting values

$156.3 = 31\times t + 0$

t = $\frac{156.3}{31 }$

= 5.042 s

Similary, for the case of landing

t = 5.042 s; initial velocity, u =0

acceleration = acceleration due to gravity, g = 9.81 $m/s^2$

Substituting in $h = ut + \frac{1}{2}gt^2$

$h = 0 + \frac{1}{2} \times 9.81 \times (5.042)^2$

h = 124.694 m

So height of ramp = 124.694 m