A particle is projected from a point O on horizontal ground with an initial speed U m/s at an angle θ to the horizontal. It passes through the point (40,10). Given that tan θ = 3/4, find U, velocity c

Here is the solution to the projectile motion problem:

Question 3:
A particle is projected from a point O on horizontal ground with an initial speed $U$ m/s at an angle $\theta$ to the horizontal. It passes through the point $(40,10)$. Given that $\tan \theta = 3/4$, find:

a) The value of $U$.

b) The horizontal and the vertical components of the velocity when $t = 2$ s. State, with a reason, whether the particle is rising or falling at this instance.

c) The total time for which the particle is at a height greater than 5 m.
(take $g=10 \text{ m/s}^2$)

First, let's determine $\sin \theta$ and $\cos \theta$ from $\tan \theta = 3/4$.
Using a right-angled triangle with opposite side 3 and adjacent side 4, the hypotenuse is $\sqrt{3^2+4^2} = \sqrt{9+16} = \sqrt{25} = 5$.
So, $\sin \theta = \frac{3}{5}$ and $\cos \theta = \frac{4}{5}$.

The equations of motion for a projectile are:
Horizontal displacement: $x = (U \cos \theta) t$
Vertical displacement: $y = (U \sin \theta) t - \frac{1}{2} g t^2$

a) The value of $U$.
The particle passes through the point $(x,y) = (40,10)$.
Substitute $x=40$, $y=10$, $\cos \theta = 4/5$, $\sin \theta = 3/5$, and $g=10 \text{ m/s}^2$ into the equations: