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
Best Answer
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 $g=9.81\text{ms}^{-2}$.