You are playing paintball and realize that you ran use your newfound knowledge o

Question

You are playing paintball and realize that you ran use your newfound knowledge of physics to make a really epic shot. Your friend is standing on the roof of a 51.8 in building behind a 67.3 m thin billboard and thinks he's safe from getting hit by you on the ground below (see figure). Your paint gun shoots paint balls at a speed of 42 m/s, and you hold it at 1.5 in above the ground when you fire. a. If the paintball just barely goes over the billboard (in other words the highest point of it's trajectory is 67.5 in from the ground), what angle did you shoot it at? b. What was your horizontal distance from the billboard? c. How long does it take to rise to its highest point just above the billboard? d. Assuming your friend managed to run and get out of the way. how far from the edge of the roof does the paintball hit the building?

Explanation / Answer

part a:

initial speed=42 m/s

let angle be theta.

maximum height=67.5 m

maximum height from shooting position=67.5-1.5=66 m

for a projectile motion, maximum height is given by

v^2*sin^2(theta)/(2*g)

==>66=42^2*sin^2(theta)/(2*9.8)

==>theta=58.91 degrees

part b:

maximum height is reached in time v*sin(theta)/g=42*sin(58.91)/9.8=3.67 seconds

horizontal speed=42*cos(58.91)=21.688 m/s

then horizontal distance from the billboard=horizontal speed*time
=79.5976 m

part c:

time taken =3.67 seconds

part d:
vertical height =51.8 m

vertical displacement from shooting position=51.8-1.5=50.3 m

if time taken is t seconds

then using the formula:

displacement=initial velocity*time+0.5*acceleration*time^2

==>50.3=42*sin(58.91)*t-0.5*9.8*t^2

==>4.9*t^2-35.967*t+50.3=0

solving for t, we get t=5.46 seconds

horizontal distance travelled by that time=42*cos(58.91)*5.46=118.417 m

so distance from the edge of the building=118.417-distance of the bill board from your position-distance between billbord and the building

=118.417-79.5976-20=18.8195 m

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