You are asked to find the initial velocity of a steel ball from a spring loaded

Question

You are asked to find the initial velocity of a steel ball from a spring loaded launcher. You place the launcher on the edge of a table 1.0 meter above the floor and launch the projectile several times horizontally. If you find that the average distance the ball travels is 2.8 meters, what is the initial velocity of the ball? To solve a problem like this, use equations for the y-motion to find the time and then the x-motion to find the velocity. time = 45 s velocity = 6.2 m/s 0/2 You now place the launcher on the floor and shoot a steel ball at an angle of 45 degree. If you had previously measured an initial velocity of 3.6 m/s for this launcher, then how far from the launcher do you expect the ball to hit the floor? time = distance =

Explanation / Answer

as ball comes back to floor,

vertical displacement, y = final vertical position - initial vertical position

y = 0 - 0 = 0

initial vertical velocity, v0y = 3.6 sin45 = 2.55 m/s

ay = - 9.8 m/s^2

Applying y= v0y t + ay t^2 / 2

0 = 2.55 t - 9.8 t^2 /2

t = 0 or 0.52 sec

hence time = 0.52 sec ............Ans

in horizontal, there is no acc so horizontal velocity will remain same.

x = v0x t

= (3.6 cos45) (0.52)

= 1.32 m ........Ans

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