Question

A projectile is to be thrown horizontally from the top of a tower 1.7m high. Calculate its initial velocity of projection if it hits perpendicularly an incline of 37 starting from foot of tower

Answer 1

Answer:

$\sqrt{2}$

Explanation:

root 2h/g =√2*1.7/10=34

Answer 2

Example:-

A projectile is thrown horizontally from top of a tower strikes the ground after 3sec at an angle 450 with horizontal.find height of tower and speed which body was projected.{g=9.8m/s2}

Answer :-

The projectile is thrown horizontally from the top of the tower, therefore the initial velocity in the vertical direction is zero.

Initial velocity in vertical direction, uy = 0

The time taken by the projectile to reach the ground, t = 3 s

​Using the equation of motion in the vertical direction, the height of the tower is given as:

h = uyt+0.5gt2 = 0 + 0.5×10×32 = 45 m

​Let us find out the velocity of the projectile in the vertical direction, when the projectile hits the ground.

vy = uy +gt = 0 +10×3 = 30 m/s

This velocity is the vertical component of the final velocity of the projectile:

vsinθ​ = 30

⇒​vsin45o = 30

⇒v2√=30⇒v=302√ m/s

The final velocity in the horizontal direction:

vx=vcos45o=302√×12√=30 m/s

Applying equation of motion once again in the horizontal direction:

vx=ux+axt⇒ux=vx−axt=30−0=30 m/s