Question

A 67.5 kg sprinter exerts a force of 775 N on the starting block which makes a 20 degree angle to the ground.

a) What was the horizontal acceleration of the sprinter?
b) If the force was exerted for 0.340s, with what speed did the sprinter leave the starting block?
horizontal force component = 775 cos 20
so
a = F/m = (775/67.5)cos 220 meters/ second^2

force*time = impulse = momentum change = m v
so
775 cos 20 * 0.340 = 67.5 v

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