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A particle of mass $2\;kg$ is fired up a smooth slope of length $4 \;m$, with initial speed $10\;m/s$, inclined at an angle $30^\circ$ degrees above horizontal. What is the speed of the particle at top of the slope?

I tried using $mg\sin30^\circ$ to find the net force and then the acceleration of the particle using Newton’s second law. After that I substituted initial velocity of $10\;m/s$, acceleration of $5\;m/s^2$ and the distance of $4 \;m$ into the equation $v^2 = u^2 + 2as$ to find the velocity at top of the slope. Have I missed the fact that the motion is not in a straight line so the equation would not work?

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    $\begingroup$ How could it not work, if you tried it correctly. Start with what is the vertical height reached by the particle? What is the particles increase in gravitational potential energy at that height? $\endgroup$
    – Bob D
    Commented Apr 9, 2021 at 18:59
  • $\begingroup$ Vertical height reached by the particle is 2 metres using 4sin30 $\endgroup$
    – Anay Chadha
    Commented Apr 9, 2021 at 19:00
  • $\begingroup$ OK, then what is the increase in gravitational potential energy? $\endgroup$
    – Bob D
    Commented Apr 9, 2021 at 19:01
  • $\begingroup$ 2*10*2 = 40J then setting 40J = 1/2(2)v^2 would give v as sqrt(40) $\endgroup$
    – Anay Chadha
    Commented Apr 9, 2021 at 19:02
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    $\begingroup$ No, no. For conservation of energy 40 J is the decrease in kinetic energy at the top not equal to the kinetic energy at the top. Now do you know what to do? $\endgroup$
    – Bob D
    Commented Apr 9, 2021 at 19:06

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The acceleration used in the equation $v^2=u^2+2as$ should be $-g\sin 30^\circ\approx-5m/s^2$ so that $v^2=10^2-2\cdot5\cdot4$ so that $v=\sqrt{60}=2\sqrt{15}\;m/s$. The kinematic equation can be recognized as the energy conservation equation in disguise.

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Its much better to use energy conservation for this type of sums.You probably didn't take care of the fact that acceleration is not positive but negative. Also motion might not be along a horizontal line but its certainly along a straight line.

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