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}

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=30v=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+axtux=vx-axt=30-0=30 m/s

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