Question

A motorcycle stunt rider moving horizontally takes off from point 1.25m above the ground landing 10m away. What was the take off speed?

Answer 1

Answer:

20

Explanation:

range=vi^2/2g

Answer 2

Concept:

Projectile motion is a type of motion that an item or particle (a projectile) experiences when it is projected near the Earth's surface and moves along a curved route only due to gravity (in particular, the effects of air resistance are passive and assumed to be negligible).

Find:

The take-off speed of the motorcycle.

Solution:

Since the motorcycle takes off horizontally, its initial vertical velocity is zero.

Vertical distance s $=1.25m$

$s=ut+\frac{1}{2}at^2\\1.25=0(t)+\frac{1}{2}*9.8*t^2\\t=0.505sec$

There is no acceleration in the horizontal direction, so the motorcycle moves at a constant speed.

$speed=\frac{distance}{time} \\\\speed=\frac{10}{0.505}\\\\ speed=20ms^{-1}$

Hence, the take-off speed is $20ms^{-1}$.

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