Question

A 0.13 kg ball of dough is thrown straight up into the air with an initial speed of 18 m/s. The acceleration of gravity is 9.81 m/s2 .What is its momentum halfway to its max- imum height on the way up?

Answer 1

Answer:

1.65 kg.m/s

Explanation:

mass of ball (m) = 0.13 kg

initial speed (u) = 18 m/s

acceleration due to gravity (g) = 9.81 m/s^{2}

angle of projection (p) = 90°

to find the momentum half way to its maximum height we first have to know the maximum height, and the velocity at the maximum height.

maximum height (h) = $\frac{u^{2}. (sinp)^{2} }{2g}$

maximum height (h) = $\frac{18^{2}. (sin90)^{2} }{2x9.8}$ = 16.53 m

velocity at half the maximum height (v) can be gotten from the equation of motion

$v^{2} =u^{2} -2a.\frac{h}{2})$ ( the negative sign is because the ball is moving upwards)

$v^{2} =18^{2} -(2x9.8x\frac{16.53}{2})$

$v^{2} =324 - (161.99)$

$v^{2} =162.01$

v = $\sqrt{162.01}$

v= 12.72 m/s

momentum of the ball half way to its maximum height = mass x velocity = 0.13 x 12.72 = 1.65 kg.m/s