Question

If he leaves the ramp with a speed of 35.0 m/s and has a speed of 33.0 m/s at the top of his trajectory, determine his maximum height (h) (in m) above the end of the ramp. Ignore friction and air resistance.

Answer 1

Answer:

H = 6.93 m

Explanation:

given data

velocity v = 35 m/s

horizontal component Vx = 33 m/s

solution

we get here maximum height so first we get vertical component here that is express as

Vy = $\sqrt{v^2- Vx^2}$        .........................1

put here value

Vy = $\sqrt{35^2- 33^2}$

Vy = 11.66 m/s

and

now we get height

H = $\frac{Vy^2}{2g}$        .............................2

put here value

H = $\frac{11.66^2}{2\times 9.8}$

H = 6.93 m