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A small ball is projected with speed $16\;\mathrm{m/s}$ at an angle of $45^\circ$ above the horizontal from a point on the horizontal ground. Calculate the period of time before the ball lands, for which the speed of the ball is less than $12\;\mathrm{m/s}$. ($g=10\;\mathrm{m/s^2}$)

I have calculated the answer to be 0.8 seconds. Then I checked the “mark scheme” and the example solution was:

$$v^2 = 12^2-(16\cos(45))^2 \implies v=4$$ $$-4=4-gt \implies t=0.8 s$$

I could not figure out what exactly they were doing. What am I missing?

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  • $\begingroup$ I'm confused, didn't you and the book get the same answer? $\endgroup$
    – JMac
    Commented Apr 19, 2017 at 11:49
  • $\begingroup$ Yes, but the way presented in the mark scheme, I do not see they arrived at the answer. $\endgroup$
    – blackened
    Commented Apr 19, 2017 at 11:51
  • $\begingroup$ They seem to be using Pythagoras and then one of the four kinematic equations. What was your approach to get the same answer? $\endgroup$
    – Steeven
    Commented Apr 19, 2017 at 12:02

1 Answer 1

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Its something like this:

Let $v$ be the vertical component of the velocity of the ball when its net velocity is 12m/s. Then $$ v^2 + (16cos(45))^2 = 12^2$$ because the horizontal component of velocity is unchanged ($=16cos(45)$).

Finally, apply the familiar $$v = u + at$$ equation to compute the time for which the vertical component of velocity of the ball is less than $v$. This will be the time during which the vertical velocity changes from $+v$ to $-v$. (Keep in mind that the vertical component has initially a very large positive value, it slows down, reaches $+v$, goes to zero, then starts increasing in negative magnitude during it which it reaches $-v$ and then starts becoming even mrore negative.)

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  • $\begingroup$ Okay. I thought they would denote the vertical component with $v_y$ or $\dot{y}$. $\endgroup$
    – blackened
    Commented Apr 19, 2017 at 12:12
  • $\begingroup$ Yeah, Its not much of a "solution". They haven't explained anything. $\endgroup$
    – Abhijeet Melkani
    Commented Apr 19, 2017 at 12:13

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