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A projectile is to be fired in a such a way that it will return to the car after it has travelled 100m. The speed of the projectile must be (ignore air drag)

MY SOLUTION

u means initial velocity while R means range and T is time of flight. $\theta$ is the angle made by the initial velocity with the horizontal.

What the question basically means is, the projectile is fired from the moving car, which goes forward while the car also moves forward, and both the projectile and car coincide at a point 100m from the point of projection.

Obviously, the time of flight of the projectile is 4 seconds and it’s range will be 100m $$4=\frac{2u\sin\theta}{10}$$ So $u\sin\theta=20$ and $$100=\frac{u^2\sin2\theta}{10}$$ Using value of $u\sin\theta$ $$u\cos\theta=25$$ So u will be $\sqrt{u_x^2 + u_y^2}$

$=32.0156$

Sine the car is moving with velocity 25, the velocity of the projectile must be fired with velocity 7.0156m/s

The given answer is 19.6m/s. Can anyone please verify that? Thanks a lot

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  • $\begingroup$ Can you add the definitions of $u$ and $\theta$ ? $\endgroup$ – Evargalo Aug 27 '19 at 14:08
  • $\begingroup$ If the projectile is to be fired from the moving car, then solve the problem in the car’s reference frame. The projectile’s motion looks particular simple there. $\endgroup$ – amd Aug 27 '19 at 14:17
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To a stationary observer, the projectile is launched at velocity $u$ at launch angle $\theta$ to the horizontal, using your notation.

Using your formula for time taken to travel the entire range, $t=\frac {2u}g\sin\theta$, and putting $t=4$ gives $u\sin\theta=2g$.

The horizontal component, $u\cos\theta$, is "supplied" by the moving car, i.e, equal to the velocity of the car, i.e. $25\text{ms}^{-1}$.

The vertical component, $u\sin\theta$, is the velocity at which the projectile is launched vertically upwards in the moving car. This has already been computed above as $2g$, which is $19.6\text{ms}^{-1}$, at $g=9.81\text{ms}^{-2}$.

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Their solution is correct (for the y component of velocity):

The motion in the y coordinate satisfies:

$y = y_0 + v_0 t -\frac{1}{2}g t^2$, with $y_0 =0$.

As you’ve said, you know $y(4)=0$, so:

$0 = (v_0 -\frac{1}{2}g t) t$

Which to have a solution at $t=4$, means that:

$0 = (v_0 -\frac{1}{2}g 4)$.

So $v_0 = 2(9.8)$m/s

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  • $\begingroup$ Could you solve it the way I did? I know mines way longer, but that would help me get some insight on where i went wrong. $\endgroup$ – Aditya Aug 27 '19 at 14:19
  • $\begingroup$ I don’t know. How did you get your first equation, and what are $u$ and $\theta$? $\endgroup$ – Joe Aug 27 '19 at 14:23
  • $\begingroup$ I added that in the edit. $\endgroup$ – Aditya Aug 27 '19 at 14:47
  • $\begingroup$ Ok, I agree with all your steps, except where you subtract 25m/s. The speed the projectile must be fired is the 32.015... $\endgroup$ – Joe Aug 27 '19 at 14:59
  • $\begingroup$ But the book’s answer is the vertical component of the velocity. $\endgroup$ – Joe Aug 27 '19 at 15:00
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Your calculation is all correct, except that the velocity you obtained in your answer (32 m/s) is relative to the ground. However, the question asks for the velocity relatively to the car, which is natural since the projectile is fired from the car. So, you need to remove the horizontal velocity $u_x = 25 m/s$ of the projectile in your answer, which happens to be the same as the car.

Thus, the answer is just $u_y = 20 m/s$, because you used $g\approx 10$, as opposed to 9.8. Note that, relatively to the car, the projectile is shot vertically into the air and returns to the car vertically. So, moving with the car, you don't need to worry about any horizontal velocity.

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  • $\begingroup$ But I subtracted the car’s velocity from my answer, so it’s still relative to the car right? My final answer is 7 $\endgroup$ – Aditya Aug 27 '19 at 14:48
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    $\begingroup$ Relatively to the ground, the projectile is fired at an angle, with $u_x=25m/s$ and $u_y=20m/s$. Since the car is moving horizontally, you need to subtract car's velocity from its horizontal component $u_x$, not from its total velocity $\sqrt{u_x^2+u_y^2}=32m/s$ $\endgroup$ – Quanto Aug 27 '19 at 15:14
  • $\begingroup$ So basically 25 and 25 get subtracted and only 20 remains $\endgroup$ – Aditya Aug 27 '19 at 15:34
  • $\begingroup$ Correct. Horizontal velocity does not play any role. $\endgroup$ – Quanto Aug 27 '19 at 15:37
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you seem to take $$g=10m/s^2$$? then your first equation gives $$u_y=2u*sin\Theta)=20m/s$$ Theta being the angel seen from outside. this is the only velocity one has to give the projectile seen from the car from outside id would be $$v=\sqrt{25^2+20^2}m/s=32.015$$ the only error you made is evaluating $$\sqrt{u_x^2 + u_y^2} for u_y $$ 25^2+u_y^2=32,0156^2 ==> u_y^2 =1025-625=400

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