A projectile of mass 0.994 kg is shot from a cannon The end of the cannon\'s bar

Question

A projectile of mass 0.994 kg is shot from a cannon The end of the cannon's barrel 1S at height 6.5 m, as shown in the figure. The initial velocity of the projectile is 9.5 m/s The end of the cann e projectile rises to a maximum height of Ay above the end of the cannon's barrel and strikes the ground a horizontal distance Ar past the end of the cannon's barrel Determine the vertical component of the initial velocity at the end of the cannon's bar- rel, where the projectile began its trajectory. The acceleration of gravity is 9.8 m/s 2 Answer in units of m/s.

Explanation / Answer

a)The vertical component of intial velocity will be:

vy = v0 sin(theta)

vy = 9.5 x sin46 = 6.83 m/s

Hence, vy = 6.83 m/s

b)The maximum height will be given by: (using v^2 = u^2 + 2 a S)

Hmax = vy^2/2g + h

where h is the barrel height

Hmax = 6.83^2/2 x 9.8 + 6.5 = 8.88 m

Hence, Hmax = 8.88 m = 8.9 m = 9 m.

c)time to reach max height will be given by: (using v = u + at)

t = vy/g = 6.83/9.8 = 0.697 s

Hence, t = 0.697 s = 0.7 s

d)

from eqn of motion

s = ut + 1/2 gt^2

-6.5 = 6.83 t - 1/2 x 9.8 t^2

4.9 t^2 - 6.83 t - 6.5 = 0

the above quadratic eqn gives us

t = 2.04 sec

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