Question

A long jumper leaves the ground at an angle of 25 degrees with respect to the horizontal with a resultant velocity of 30 ft/s. What was the horizontal velocity of the jumper at takeoff? What was the vertical velocity at takeoff? How high did the COM rise above the point of takeoff?

Answer 1

Answer:

The horizontal component of the velocity is 27.19 ft/s

The vertical component of the velocity at takeoff is 12.68 ft/s

The vertical distance traveled by the jumper is 8.2 ft

Explanation:

Given;

angle of projection or takeoff, θ = 25°

the resultant velocity, v = 30 ft/s

The horizontal component of the velocity is given by;

$v_x = vcos \theta\\\\v_x = 30 * cos(25)\\\\v_x = 27.19 \ ft/s$

The vertical component of the velocity at takeoff is given by;

$v_y = vsin \theta\\\\v_y = 30 * sin(25)\\\\v_y = 12.68 \ ft/s$

The vertical distance traveled by the jumper is given by;

$v_y_f^2 = v_y_o^2 +2(-g)h\\\\v_y_f^2 = v_y_o^2 -2gh\\\\0 = 12.68^2 - (2*9.8)h\\\\0 = 160.78 - 19.6h\\\\19.6h = 160.78\\\\h = \frac{160.78}{19.6}\\\\ h = 8.2 \ ft$