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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