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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 $s$.

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

2 Answers 2

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

Which you can approximate to 7.75m/s.

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The question is asking for net magnitude of velocity . You are finding only the y component of velocity.

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