Question

A shell fired from the ground is just able to cross in the horizontal direction to the top of wall 90 M away and 45 M high. What is the direction of projection of the shell

Answer 1

The vertical component :

The final velocity = 0

Initial velocity = u

g = 10

We use the following equation of motion.

V² = u² - 2gs

0 = u² - 2 × 10 × 45

0 = u² - 900

u² = 900

u = 30

Horizontal component :

R = (u² Sin 2Ф) / g

Substituting u in thus we have :

90 = (30² Sin 2Ф) / 10

900 = 900 Sin 2Ф

Sin 2Ф = 1

Sin ⁻¹ 1 = 90

2Ф = 90

Ф = 45°

Answer 2

Answer:

theta will be 45 degree