You need to calibrate your giant paintball bazooka, so you're fired at an angle of 60° and you find that the giant paintball splatter the ground at a horizontal distance of 107m from the end of the muscle in a distance of 1.7 M below the end of the muzzle. What is the muzzle velocity, V0, of your giant paint ball bazooka?

Vx = V0 cos 60 = V0 ./ 2        horizontal speed

Vy = V0 sin 69 = V0 * .866 = .866 V0   vertical speed

t = 107 / Vx = 214 / V0   time to travel 107 m

h = Vy t - 1/2 * g t*2   time to fall 1.7 m

-1.7 = .866 V0 * 214 / V0 - 4.9 * (214 / V0)^2           substituting

224400 / V0^2 = 187

V0 = (224400 / 187)^1/2 = 2=34.64 m/s

t = 214 / 34.64 = 6.178 stc

Check:    h = Vy t - 1/2 g t^2

-1.7 = .866 * 34.64 * 6.178 - 4.9 * 6.178^2

-1.7  = -1.69

what was the acceleration of the 2 kg cart?

⇒ 16.0 cm/s2

what was the acceleration of the 4 kg cart?

⇒ 8.1 cm/s2

what was the acceleration of the 6 kg cart?

⇒ 5.3 cm/s2

b.a device that inputs heat into a cool object