Question

5. Initial velocity of a projcetile is (3î + 4ſ) m/s.Find its
maximum height and horizontal range.(g = 10 m/s)
(2015)​

Answer 1

It is given that the initial velocity of the projectile is ( 3 î + 4 j^ )m/s.

So, initial vertical velocity = 4m/s and initial horizontal velocity = 3 m/s.

- At the highest point, vertical velocity becomes zero. Since acceleration is constant (g), we can apply the equations of motion over here. So we can use the equation v^2 = u^2 + 2aS to find the maximum height.

- Once the particle reaches the ground, its net vertical displacement is zero. So, we can use S=ut+(1/2)at^2 to find the time of flight. There is no horizontal acceleration so the range (horizontal distance travelled) is simply the distance travelled by the horizontal component of the velocity in the total time of flight.

I hope these hints were helpful.