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Question

A projectile of mass 0.929 kg is shot from a cannon. The end of the cannon's barrel is at height 6.8 m. The initial velocity of the projectile is 9.5 m/s. The projectile rises to a max height of delta y above the end of the cannon's barrel and strikes the ground a horizontal distance delta x past the end of the cannon's barrel. Gravity is 9.8 m/s^2

The vertical component of the initial velocity at the end of the barrel where the projectile begins it's trajectory is 7.58 m/s and the max height delta y the projective achieves after leaving the end of the cannon's barrel is 2.93 m.
A) Determine the time it takes the projectile to reach its in max height in seconds.
B) How long does it take the projectile to hit the ground in seconds?
C) Find the magnitude of the velocity vector when the projectile hits the ground. Answer in m/s
D) Find the magnitude of the angle(with respect to the horizontal) the projectile makes when impacting the ground. Answer in degrees.
E) Find the range delta x

Answer 1

vy(t)=7.58-9.8*t
a) when the projectile reaches max height, the vy=0
solve for t
t=0.773 seconds

b) when y(t) =0, the projectile strikes the ground

0=6.8+7.58*t-.5*9.8*t^2
Solve for t, use the positive root
t=2.182 seconds

c)
vy(2.182)=7.58-9.8*2.182
vy=-13.8

vx can be found from the initial velocities as
vx=sqrt(9.5^2-7.58^2)
vx=5.73

the magnitude is
sqrt(13.8^2+5.73^2)
14.9 m/s

d)
Tan(θ)=13.8/5.73
67.45 degrees below horizontal

e)
x(2.182)=5.73*2.182
12.5 meters

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