a projectile is thrown horizontally from top of a tower strikes the ground after 3sec at an angle 45^{0} with horizontal.find height of tower and speed which body was projected.{g=9.8m/s^{2}}

Initial velocity in vertical direction, u

_{y}= 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 = u

_{y}t+0.5gt

^{2}= 0 + 0.5×10×3

^{2}= 45 m

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

v

_{y}= u

_{y}+gt = 0 +10×3 = 30 m/s

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

vsinθ = 30

⇒vsin45

^{o}= 30

$\Rightarrow \frac{v}{\sqrt{2}}=30\phantom{\rule{0ex}{0ex}}\Rightarrow v=30\sqrt{2}\text{m/s}$

The final velocity in the horizontal direction:

${v}_{x}=v\mathrm{cos}{45}^{o}=30\sqrt{2}\times \frac{1}{\sqrt{2}}=30\text{m/s}$

Applying equation of motion once again in the horizontal direction:

${v}_{x}={u}_{x}+{a}_{x}t\phantom{\rule{0ex}{0ex}}\Rightarrow {u}_{x}={v}_{x}-{a}_{x}t=30-0=30\text{m/s}$

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