a.) An electron is ejected horizontally at a speed of 1.6×10 6 m/s from the elec

Question

a.) An electron is ejected horizontally at a speed of 1.6×106 m/s from the electron gun of a computer monitor.

If the viewing screen is 39 cm from the end of the gun, how far will the electron travel in the vertical direction before hitting the screen?

Express your answer using two significant figures.

b.) An artillery crew wants to shell a position on level ground 22 km away.

If the gun has a muzzle velocity of 830 m/s , to what angle of elevation should the gun be raised?

Express your answer using two significant figures.

c.) A ball rolls horizontally with a speed of 8.9 m/s off the edge of a tall platform.

If the ball lands 7.2 m from the point on the ground directly below the edge of the platform, what is the height of the platform?

Express your answer using two significant figures.

Explanation / Answer

s = vt
time taken to reach screen = s/v
t=0.39/1.6 x 10^6
t =2.437 x 10^-7 sec

Distance fallen due to gravity in this time will be
s = 0.5gt^2
s = 0.5 x 9.81 x (2.437 x10^-7)^2
s =2.9133 x 10^-13 m
s = 0.2913pm

b..Vo = 830 m/s
R = 22 m
g = 9.80665 m/s^2 (standard definition)

Range equation

R = [ (Vo^2) * (sin2) ] / g

Solve for angle gives, then solve with values

sin2 = [ R g ] / (Vo)^2

sin2 = [ (22000 m) * (9.80665 m/s^2) ] / (830 m)^2

sin2 =0.3131
2 = 18.2506
=9.1253°

c.

find t
x=vt
8.9m=7.2t
t= 1.23611s

the time it land 8.9m away from the base is the time it takes the ball to fall.

x=1/2at^2
x=(1/2)(-9.8m/s^2)(1.23611s)^2
x= -7.487m

The platform is 7.487m high

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