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?

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;

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

The vertical distance traveled by the jumper is given by;

i would say the answer is food.

hope this !

it has to be b, about 97 percent.