Question

A projectile of mass 0.99 kg is shot from a
cannon. The end of the cannon’s barrel is
at height 6.6 m, as shown in the figure. The
initial velocity of the projectile is 9.2 m/s.
The projectile rises to a maximum height of
∆y above the end of the cannon’s barrel and
strikes the ground a horizontal distance ∆x
past the end of the cannon’s barrel.
Find the time it takes for the projectile to
reach its maximum height.
The acceleration
of gravity is 9.8 m/s
Answer in units of s



the angle given is 47◦




b)How long does it take the projectile to hit the
ground?
Answer in units of s

c)Find the range ∆x of the projectile.
Answer in units of m

Answer 1

Vo = 9.2m/s[47o]
Xo = 9.2*Cos47 = 6.27 m/s.
Yo = 9.2*sin47 = 6.73 m/s.

a. Y = Yo + g*Tr = 0 @ max ht.
Tr = -Yo/g = -6.73/-9.8 = 0.687 s. =
Rise time or time to reach max ht.

b. h max = ho + Yo*Tr + 0.5g*Tr^2 =
6.6 + 6.73*0.687 - 4.9*0.687^2 = 8.91 m.
Above gnd.

c. h = 0.5g*Tf^2 = 8.91 m.
4.9*Tf^2 = 8.91
Tf^2 = 1.82
TF = 1.35 s. = Fall time.

Range = Xo*(Tr+Tf)=6.27 * (0.687+1.35) =
12.77 m.