A person would like to jump over a river using a motorcycle and a ramp. the fric

Question

A person would like to jump over a river using a motorcycle and a ramp. the frictionless ramp is inclined at an angle of 49.3 degrees, has a length of 59 meters, and the motocycle's engine produces a constant force of 704 newtons directed up the ramp, parallel to the ramp. the width of the river is 262 ,meters, and the landing on the other side of the river is at the same height at the end of the ramp. if the motorcycle is to just barely make it across the river and it's total mass is 58.1 kg, what is should the magnitude of the motorcycle's velocity be at the bottom of the ramp in m/s?

Explanation / Answer

let the velocity with which it leaves the surface be v, angle with which it leaves will be theta

Range of the projectile motion = width of the river

V^2 sin(2 * theta) / g = 262

=> v^2 = 262 * 9.81 / sin(98.6) = 2570.22 / 0.988756 = 2599.45

=> v = 50.98 m / sec

let the velocity at the bottom be u

From Equation of linear motion

v^2 - u^2 = 2 * a * s

Now, we have to find the acceleration a

writing free body diagram on bike we get

F - mg*sin(theta) = m*a

704 - 58.1 * 9.81 * sin(49.3) = 58.1 * a

704 - 432.107 = 58.1 * a

=> a = 4.6797 m/s^2

From, v^2 - u^2 = 2 * a * s

(50.98)^2 - u^2 = 2 * 4.6797 * 59

=> 2598.96 - u^2 = 552.2046

u^2 = 2046.7558

=> u = 45.24 m / sec

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