A projectile of mass 0.342 kg is shot from a cannon. The end of the cannon’s bar
ID: 2059281 • Letter: A
Question
A projectile of mass 0.342 kg is shot froma cannon. The end of the cannon’s barrel
is at height 6.3 m, as shown in the figure.
The initial velocity vi of the projectile has a
horizontal component of 7.3 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.
Determine the time it takes for the pro-
jectile to reach its maximum height and how long it takes to hit the ground. The
acceleration of gravity is 9.8 m/s2 .
Answer in units of s
Explanation / Answer
conservation of energy:
(1/2)m(vi)^2 = mgy
(vi)^2 =2gy
let the y component of initial velocity be viy
(vi)^2 = (vix)^2 + (viy)^2
(2gy)^2 = 7.3^2 + (viy)^2
viy= sqrt[(19.6y)^2 - 53.29]
let time it takes to reach maximum be tmax, when vy is 0.
a=[(vy)-(viy)]/(tmax)
-9.8=(0-viy)/(tmax)
tmax = sqrt[(19.6y)^2 - 53.29] / 9.8 sec
time it takes to hit the ground is the distance over x-component velocity, since it is constant.
tend = x/7.3 sec