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, 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_yt+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.
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
$$\begin{gathered} \Rightarrow \frac{v}{\sqrt{2}}=30 \\ \Rightarrow v=30\sqrt{2} \text{m/s} \end{gathered}$$
The final velocity in the horizontal direction:
$v_x=v\cos {45}^o=30\sqrt{2}\times \frac{1}{\sqrt{2}}=30 \text{m/s}$
Applying equation of motion once again in the horizontal direction:
$$\begin{gathered} v_x=u_x+a_{x}t \\ \Rightarrow u_x=v_x-a_{x}t=30-0=30 \text{m/s} \end{gathered}$$