Question

A daredevil on a motorcycle leaves the end of a ramp with a speed of 35.0 m/s .If his speed is 33.0 m/s when he reaches the peak of the path, what is the maximum height that he reaches? Ignore friction and air resistance.

Answer 1

Let the angle at which the motorcycle leaves the ramp makes an angle $\theta$ with the horizontal

the horizontal component= $35cos\theta$

vertical component = $35sin\theta$

At the maximum height the vertical component of the velocity becomes zero

$35cos\theta=33 \\cos\theta=\dfrac{33}{35}$ $sin\theta=\dfrac{11.67}{35}$

vertical component =$35sin\theta =11.67m/s$

at maximum height vertical component of velocity is zero

So applying formula

$v^2=u^2+2gS$

$0=(11.67)^2+2(-10)S \\S=136/20=6.8 m$