While on avanlanche patrol in the mountains, you stand at the base of a long str

Question

While on avanlanche patrol in the mountains, you stand at the base of a long straight 16.1° slope and shoot a projectile towards it, angle 60° above horizontal. The projectile reaches its highest point 3.00 seconds later.

a) What was the projectile's speed as it exits the muzzle of the launcher?

b) How far above you ( vertically ) does the projectile get?

c) How far from you horizontally is the projectile at its highest point?

d) How long after firing dows the projectile land on the sloped incline?

Explanation / Answer

let speed at which it exits the muzzle is v m/s

part a:

time to reach maximum height=v*sin(theta)/g

=v*sin(60)/9.8

given this time period is 3 seconds,

v*sin(60)/9.8=3

==>v=33.9482 m/s

part b:

maximum height=v^2*sin^2(theta)/(2*g)

=33.9482^2*sin^2(60)/(2*9.8)

=44.1 m

part c:

horizontal distance=horizontal speed*time to reach maximum height

=v*cos(60)*3

=50.9223 m

part d:

velocity along perpendicular to the inclince=v*sin(60-16.1)=23.5397 m/s

acceleration along perpendicular to the inclince=-9.8*cos(16.1)=-9.4156 m/s^2

as displacement along perpendicular to the incline 0 when the projectile hits the sloped inclince,

using the formula:

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

==>0=23.5397*t-0.5*9.4156*t^2

==>t=23.5397/(0.5*9.4156)=5 seconds

so the projectile will land on the sloped incline at 5 seconds.

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