A projectile of mass 0.342 kg is shot from a cannon. The end of the cannon’s bar

Question

A projectile of mass 0.342 kg is shot from
a 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

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