A ball is thrown horizontally from a cliff such that it strikes ground after 5 secs the line of sight from the point of projection to the point of hitting makes an angle 37° with horizontal what is initial velocity of projection

i just want the answer

To find the initial velocity of projection, we need to analyze the components of motion separately in the horizontal and vertical directions.

Let's denote the initial velocity as "v₀" and the time of flight as "t".

In the horizontal direction:
Since the ball is thrown horizontally, the initial horizontal velocity is constant throughout the motion.

Horizontal velocity (vₓ) = v₀

In the vertical direction:
We can consider the motion of the ball using the equations of linear motion.

Vertical displacement (h) = 0 (since it starts and ends at the same vertical level)
Vertical acceleration (a) = -9.8 m/s² (acceleration due to gravity, acting downwards)
Time (t) = 5 s
Angle of projection (θ) = 37° (with the horizontal)

Using the equation of motion: h = v₀yt + 0.5at²

Since h = 0, we can rewrite the equation as:
0 = v₀sin(θ)t + 0.5at²

Now we can substitute the known values:
0 = v₀sin(37°)(5s) + 0.5(-9.8 m/s²)(5s)²

Simplifying the equation gives us:
0 = 5v₀sin(37°) - 122.5

Rearranging the equation to solve for v₀ gives us:
v₀sin(37°) = 122.5
v₀ = 122.5 / sin(37°)

Using the sine function, we can find the value of sin(37°) by taking the inverse sine (also known as arcsin) of 37°:
sin⁻¹(37°) = 0.6018

Substituting this value back into the equation gives us:
v₀ = 122.5 / 0.6018

Evaluating the equation:
v₀ ≈ 203.60 m/s

Therefore, the initial velocity of projection is approximately 203.60 m/s.

what distance does it fall in 5 seconds?

h=1/2 *9.8*25
how far horizontally did it go?
tan 37=distanc/h
distance= h*tan37
velocity= distance/5sec