Physics Acceleration

Question

An athlete accelerates to 10 m/sec and then jumps at a 25° angle. How far will he jump?

Answer 1

Solution:

The initial velocity of the athlete at the point of the jump is V_o = 10 m/s

The distance that he will travel at ∝=25° is the horizontal component of a projectile with V_o = 10 m/s

x = V_o cos (∝) t . So you need to find the time t to solve this

Find t from the vertical component y = -1/2gt² + V₀sin∝t + y_o where y = y_o = 0, the explanation to that is when he was still on the ground at the point he jumps, his initial altitude y_o=0 , and when he lands after the jump his final altitude is y =0. Substituting the above:

0 = -1/2 gt² + V_osin∝ t + 0 , 1/2 gt² = V_osin∝ t , 1/2 gt = V₀ sin∝_, gt = 2V_osin∝, t = 2V_osin∝ / g

t = 2(10m/s)(sin 25º) / 9.81 m/s² = 0.86 s

x = 10m/s cos(25º)(0.86s) = 7.794 m, say 7.8 m