Question

A fielder tosses a 0.15 kg baseball at 26 m/s at a 36 ∘ angle to the horizontal.Part AWhat is the ball's kinetic energy at the start of its motion?Express your answer with the appropriate units.Part BWhat is the kinetic energy at the highest point of its arc?Express your answer with the appropriate units.

Answer 1

Answer:

A. 50.7 J

B. 33.18 J

Explanation:

At the start of the motion the kinetic energy of the ball would be

$E_k = \frac{mv^2}{2} = \frac{0.15*26^2}{2} = 50.7 J$

When the ball gets to the highest point, vertical velocity must be 0, only horizontal velocity contributes to the kinetic energy. Since air resistant can be neglected, horizontal energy is preserved since the start

$v_h = vcos(36^o) = 26*0.81 = 21.03 m/s$

So the kinetic energy at this point is

$E = \frac{mv_h^2}{2} = \frac{0.15*21.03^2}{2} = 33.18J$