Two runners start at the same point on a straight track. the first runs with constant acceleration so that he covers 98 yards in 10 seconds. the second runner waited 5 seconds and then throws a rock at his opponent's head. of the head and the rock are at the same level form the ground, what must the initial magnitude of the velocity be if the rock is to hit the head just at the 98 yard tape? give the answer in feet per second.

94.13 ft/s

Explanation:

Given:

Assume:

For the motion of the rock along the vertical during the time of flight, the rock has a constant acceleration in the vertically downward direction.

Now the rock has zero acceleration along the horizontal. This means it has a constant velocity along the horizontal during the time of flight.

On dividing equation (1) by (2), we have

Now, putting this value in equation (2), we have

Hence, the initial velocity of the rock must a magnitude of 94.13 ft/s to hit the opponent exactly at 98 yards.

15.3 cos 65 degrees trig

one person pulls 90n one way and the other has 100n pulling the other way.

so the net force = 90 - 100 = -10n,

or the ans can be 10n if the picture shows -90n and 100n instead.