Projectile Motion - what was the ball's speed when it left your hand?

Question

Suzanne has climbed a tree to a height of 6.20 m. You throw a ball vertically up to her and it is traveling at 5.00 m/s when it reaches her. What was the speed of the ball when it left your hand if you released it at a height of 1.10 m?

Thanks for any help!

Answer 1

Ok, let's solve the problem :

Make a graphic with Suzanne at the top of the tree, le's supposed she is on the top at 6.2 meters over the ground, and let's draw you throwing the ball.

Let's use the conservation of energy :

The ball was released at 1.10 over the ground with some speed, and will reach an altitude 6.20 at 5 m/s

then :

m*g*1.10 + mv^2 / 2 = m*25 / 2 + mg*6.2

initial (Potencial energy + kinetic energy ) = final ( potential energy + kinetic energy)

considering g = 9.8 m/s^2

canceling "m" :

9.8*1.10 + v^2 / 2 = 6.2*9.8 + 25 / 2

v = 11.17 m /s

It's correct, why ?, the velocity at the beginning must be higher than the velocity at the end, because, the gravity slows it down.

Note : You can solve it using other equation of cinematic, because you have the altitude : 6.2 - 1.10 = 5.1 m
And you have the initial velocity, and gravity, so, try it using those equations, for example :

Vf^2 = Vo^2 - 2gh

Hope that helped

Answer 2

enable A = the attitude above horizontal at which the ball is thrown. Angles below horizontal would be unfavorable. S = the pitching velocity. that's the fee of the fee (which additionally has a direction). Decompose the preliminary velocity: Vu is the preliminary velocity upward. Vu = S * sin(A) Vh is the horizontal velocity far off from the ledge. Vh = S * cos(A) the universal place formulation is P(t) = 0.5 * a * t^2 + Vo * t + Po the place a = acceleration Vo = velocity at t = 0 Po = place at t = 0 there's no horizontal acceleration. Use the throwing place as Po and beneficial values are far off from the ledge. So 25 = 0.5 * 0 * t^2 + Vh * t + 0 25 = Vh * t 25 / Vh = t The vertical acceleration is g, -9.8 m/s^2. this is unfavorable because of the fact this is directed downward. Use Po = 4, the preliminary height. the astounding height at time t would be 0 (on the floor). 0 = 0.5 * -9.8 * t^2 + Vu * t + 4 0 = -4.9 * t^2 + Vu * t + 4 replace the fee for t derived from Vh 0 = -4.9 * (25 / Vh)^2 + Vu * (25 / Vh) + 4 we don't choose variables interior the denominators, so multiply the two aspects by using Vh^2. 0 = -4.9 * 25^2 + 25 * Vu * Vh + 4 * Vh^2 0 = 4 * Vh^2 + 25 * Vu * Vh - 3062.5 Sustituting for Vh and Vu: 0 = 4 * (S * cos(A))^2 + 25 * (S * sin(A)) * (S * cos(A)) - 3062.5 0 = S^2 * (4 * cos(A)^2 + 25 * sin(A) * cos(A)) - 3062.5 clean up for S in terms of A: SQRT(3062.5 / (4 * cos(A)^2 + 25 * sin(A) * cos(A))) = S replace the three exciting values for A: A = -5; S = 40-one.3 m/s A =  0; S = 27.7 m/s A =  5; S = 22.3 m/s

Answer 3

Covervation of energy.

PE + KE before = PE + KE after

Initially PE = mgh (h is your hand height)

Final PE = mgh (h is her heigt in tree)

Initial KE = 1/2mv^2 (need to solve for this v)

Final KE = 1/2 mv^2 (v given)

Add everything up (notice that m cancels) and solve for initial v.

Answer 4

It is velocity, speed and the Gravitational pull of the earth.