Question

A shell is fired at an angle of 45 degrees above ground with an initial velocity of 100 m sec– 1 . It will hit the ground, assuming g = 10 m sec – 2 , after bout:

Answer 1

Given: A shell is fired at an angle ∅= 45°

           Initial Velocity $U_{Y}$= 100 m/s

           Here, g = 10m/s²

To Find: The span of time after the shell hits the ground.

Solution:

Let the time be = t sec

Therefore, The displacement $S_{Y}$ = 0

According to the equation of motion,

$S_{Y}$ = $U_{Y}t$ +$\frac{1}{2}$ $a_{y}t^{2}$ [ The projectile is towards Y axis ]

or, 0 = 100Sin45° + $\frac{1}{2}$(-g)t²

or, 5t² = 50√2

or, t² = 10√2

or, t = 3.37 sec

The shell will hit the ground after 3.37 sec.

Concept:

• Position-time relation - It is the second equation of motion that is used when the displacement of a particle is calculated, when the values of time, acceleration, and velocity are known. Where S = displacement, U= Initial velocity, t = time, a = acceleration and the formula is - $s=ut+ \frac{1}{2} at^{2}$

Answer 2

Answer:

14 sec

Explanation: