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

1.65 kg.m/s

Step-by-step 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) =
`(u^(2). (sinp)^(2) )/(2g)`

maximum height (h) =
`(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.(h)/(2))` ( the negative sign is because the ball is moving upwards)

`v^(2) =18^(2) -(2x9.8x(16.53)/(2))`

`v^(2) =324 - (161.99)`

`v^(2) =162.01`

v =
`√(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