# Finding the velocity of a projectile thrown from a raised platform at an angle [closed]

The following question has been giving me trouble for a while now, it states

A particle is thrown with speed $$10 \space \text{ms}^{-1}$$ from a point $$2\space \text{m}$$ from the ground, at an angle of $$45°$$ above the horizontal. What is the speed of the particle when it is at a height of $$4 \space \text{m}$$ for the second time in its motion?

I have tried the following,

$$v^2-u^2=2as \iff (v\sin(45))^2=(10\sin(45))^2+2(-10)(2)$$ $$v=\frac{\sqrt{10}}{\sin(45)} \implies v=4.47\space \text{ms}^{-1}$$

But I did not get the correct answer of $$v=7.75\space \text{ms}^{-1}$$, similarly I tried using the equation $$s=\frac{1}{2}at^2+ut$$ to first find the time taken and then plug it into $$v=at + u$$ to yield the velocity however this was to no avail.

Where am I going about wrong, I have a sneaking suspicion that it is to do with the value of $$s$$.

• Where am I going about wrong Please be aware that check-my-work questions are off-topic on PSE. Commented Jan 6, 2021 at 4:22
• @G.Smith my apologies I was unaware Commented Jan 6, 2021 at 4:46

$$v_y^2 = u_y^2 + 2as \iff v_y^2 = 100sin^2(45) - 40$$ $$v_y = \sqrt{10} \;and\; v_x \;remains\; 5\sqrt{2}$$ $$speed = \sqrt{v_y^2 + v_x^2} \implies speed = \sqrt{60}$$