At the instant a race began, a 55 kg sprinter was found to exert a
force of 800 N, at a 25° angle with respect to the ground, on the
starting block.

a.What was the horizontal acceleration of the sprinter?


b.If the force was exerted for 0.38 seconds, with what horizontal
speed did the sprinter leave the starting block?

Hellooo!!!


First of all we must find the horizontal projection of the force that
the sprinter exert FX :

Fx = F * cos25 = 800 * 0.9063 = 725.05 N

Fx will generate the horizontal acceleration Ax :

Ax = Fx / m = 725.05 / 55 = 13.183 m/s^2 .

This solve the part a).

If the force was exerted for 0.38 seconds, then the sprinter will be
uniformly accelerated for 0.38 seconds with an acceleration of 13.183
m/s^2 .
To find the final horizontal speed we must use the equation:

Vf = V0 + Ax * T = 0 + 13.183 * 0.38 = 5.01 m/s .

This solve part b).

I hope this helps, but always remember that if you need a
clarification you can post a request for it.

Regards.
livioflores-ga