1. In this experiment we will launch a steel ball at approximately 4 m/s at 50 d

Question

1. In this experiment we will launch a steel ball at approximately 4 m/s at 50 degrees above the horizontal direction. Assuming that this ball behaves as a projectile;

A) Find the expected intital y-component of the velocity as it leaves the launcher.

B) Find the time after launch at which the steel ball would be expected to reach its highest point.

C) Find the maximum height above the launcher that you would expect the steel ball to reach.

D) If the end of the launcher is 0.25 metres in the air, find the time it would take for the steel ball to first hit the ground after being fired.

E) Find the x-component of the initial velocity of the steel ball.

F) Find the distance that the steel ball would travel through the air before it first hits the ground.

Explanation / Answer

Given,

vi = 4 m/s ; theta = 50 deg

A)vy = vi sin(theta)

vy = 4 sin50 = 3.06 m/s

Hence, vy = 3.06 m/s

B)the time wil be:

t = vy/g = 3.06/9.8 = 0.312 s

Hence, t = .312 s

C)H = vy^2/2g

H = 3.06^2/2 x 9.8 = 0.48 m

Hence, H = 0.48 m

D)T = 2t

T = 0.624 s

E)vx = v cos(theta)

vx = 4 x cos50 = 2.57 m/s

Hence, vx = 2.57 m/s

F)D = vx T

D = 2.57 x 0.624 = 1.60 m

Hence, T = 1.60 m

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