Question

A stuntwoman is going to attempt a jump across a canyon that is 77 m wide. The ramp on the far side of the canyon is 25 m lower than the ramp from which she will leave. The takeoff ramp is built with a 15º angle from horizontal. If the stuntwoman leaves the ramp with a velocity of 28 m/s, will she make the jump? Why?

Answer 1

initial speed of the stuntman is given as

$v = 28 m/s$

angle of inclination is given as

$\theta = 15 degree$

now the components of the velocity is given as

$v_x = 28 cos15 = 27.04 m/s$

$v_y = 28 sin15 = 7.25 m/s$

here it is given that the ramp on the far side of the canyon is 25 m lower than the ramp from which she will leave.

So the displacement in vertical direction is given as

$\delta y = -25 m$

$\delta y = v_y * t + \frac{1}{2} at^2$

$-25 = 7.25 * t - \frac{1}{2}*9.8* t^2$

by solving above equation we have

$t = 3.12 s$

Now in the above interval of time the horizontal distance moved by it is given by

$d_x = v_x * t$

$d_x = 27.04 * 3.12 = 84.4 m$

since the canyon width is 77 m which is less than the horizontal distance covered by the stuntman so here we can say that stuntman will cross the canyon.