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.
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
2 answers
some of the answers is incorrect