Posted by steve on Tuesday, October 4, 2011 at 11:25am.
A projectile of mass 0.777 kg is shot from a cannon, at height 6.6 m, as shown in the ﬁgure, with an initial velocity vi having a horizontal component of 7.8m/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 vertical component of the initial
velocity at the end of the cannon’s barrel,
where the projectile begins its trajectory. The
acceleration of gravity is 9.8 m/s
2
.
Answer in units of m/s
Determine the maximum height the projectile
achieves after leaving the end of the cannon’s
barrel.
Answer in units of m
Find the magnitude of the velocity vector
when the projectile hits the ground.
Answer in units of m/s
Find the magnitude of the angle (with respect
to horizontal) the projectile makes when impacting the ground.
Answer in units of ◦
Find the range of the projectile from the time it leaves the barrel until it hits the ground.
Answer in units of m

pysics  Henry, Wednesday, October 5, 2011 at 8:39pm
1. PE = mgh = 0.777 * 9.8 * 6.6 = 50.26J.
PE = KE = 50.26,
KE = 00.5*0.777V^2 = 50.26
0.389V^2 = 50.26,
V^2 = 129.2,
Vo = 11.37m/s.
2. h = (Vf^2  Vo^2) / 2g,
h = (0  (11.37)^2) / 19.6 = 6.6m.
3. Vf^2 = Vo^2 + 2g*d,
Vf^2 = 0 + 19.6*6.6 = 129.36,
Vf = 11.37m/s.
4. tanA = Y/x = 11.37 / 7.85 = 1.4484.
A = 55.4 deg.
5. t(up) = Vf  Vo) / g,
t(up) = (o  11.37) / 9.8 = 1.16s.
Vf^2 = Vo^2 + 2g*d,
Vf^2 = 0 + 19.8*13.2 = 258.72,
Vf = 16.1m/s.,Down.
t(dn) = (Vf  Vo) / g,
t(dn) = (16.1  0) / 9.8 = 1.64s.
T = t(up) + t(dn) = 1.16 + 1.64 = 2.8s
= Time in flight.
Range = Vh*T = 7.8 * 2.8 = 21.9m.
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