Solved: A 12.0-kg shell is launched at an angle of 55.0o

Solution 110P

Step 1:

W can resolve the components of velocity as vertical and horizontal.

Vertical component of velocity, vvertical = v sin𝜽

Provided, it is making an angle, 𝜽 = 55° with the horizontal

Then, vvertical =  150 m/s × sin 55 = 150 m/s× 0.8192 = 122.87 m/s

Since it is moving against the gravitational force, we can write,

v2 - u2 = - 2 gS

Where, v - final velocity, u - initial velocity, g - acceleration due to gravity

And S - distance travelled

In vertical motion, v = 0 m/s at a point

Therefore, 0 - 122.872  m2/s2 = - 2 × 9.8 m/s2 × S

The vertical displacement, S = 15098 m2/s2 / 19.6 m/s2 = 770.26 m

Similarly, v = u - gt

0 = 122.87 m/s - 9.8 t

Rearranging, t = 122.87 m/s / 9.8 m/s2 = 12.54 s

The time to reach maximum height will be, t = 12.54 s

Therefore, the horizontal distance for explosion, H = vhorizontal t

vhorizontal = v cos𝜽

Provided, it is making an angle, 𝜽 = 55° with the horizontal

Then, vhorizontal =  150 m/s × cos 55 = 150 m/s× 0.5736 = 86.04 m/s

Therefore, the horizontal distance for explosion, H = 86.04 m/s×12.54 s = 1078.9 m