Question

A hill is 500m high. Supplies are sent to be sent across the hill using a cannon that can hurl packets at a speed of 125m/s over the hill. The canon is located at a distance of 800m from the foot of a hill and can be moved on the ground at a speed of 2m/s; so that its distance from the hill can be adjusted. What is the shortest time in which a packet can reach the ground across the hill? Take$g = 10\dfrac{m}{{{s^2}}}$.
A. 45s
B. 60s
C. 30s
D. None of these

Answer 1

First find the minimum vertical velocity component so the shortest amount of time for the packet to cross the hill. Through it, find its horizontal component. Then find the time taken by the packet using equations of linear motion, and through it, find the distance travelled by the canon in this time.

Complete step by step answer:
Here, let h be the height of the hill which is 500m. The supplies are sent with a velocity of 125m/s which is its initial velocity denoted by u. The distance d between the canon and the hill is 800m. Let the speed at which the canon is moving is denoted by v, then v=2m/s. As the canon can be moved, we have to use both the kinematic equations of linear motion for constant acceleration, as well as the equations for projectile motion.
The vertical component of the initial velocity should be the minimum for the canon to cross the hill in the shortest amount of time. Thus,

$$\Rightarrow {u_y} = \sqrt {2 \times 10 \times 500}\\\ \Rightarrow {u_y} = 100\dfrac{m}{s} \\\$$

Here, ${u_y}$is the vertical component of velocity. Let${u_x}$be the horizontal component, then

$$\Rightarrow {u_x}^2 = {u^2} - {u_y}^2 \\\ \Rightarrow {u_x}^2 = {(125)^2} - {(100)^2} \\\ \Rightarrow{u_x} = 75\dfrac{m}{s} \\\$$

Now, the time taken for the packet to reach the top of the hill is given by;

\Rightarrow t = \sqrt {\dfrac{{2 \times 500}}{{10}}} \\\ \Rightarrow t = 10s$$ Now, when the packet has reached the top of the hill, the time taken for the packet to fall on the ground would also be 10 seconds. The horizontal distance that the canon crossed in 10 seconds would be given as follows: $$x = {u_x} \times t \\\ \Rightarrow x = 75 \times 10 \\\ \Rightarrow x = 750m$$ Thus the distance z travelled by the canon would be $$z = d - x \\\ \Rightarrow z= 800 - 750\\\ \Rightarrow z= 50m$$ The time taken by the canon to move is $$t' = \dfrac{z}{v}\\\ \Rightarrow t'= \dfrac{{50}}{2} \\\ \Rightarrow t' = 25\dfrac{m}{s}$$ The total time taken by the packet would be $$T = T + 2t\\\ \Rightarrow T = 25 + 20 \\\ \therefore T = 45s$$ **Thus option A is the correct answer.** **Note:** Here, the projectile motion of the canon would not be as smooth if we take into account the air resistance, but generally in such scenarios we take that the resistance due to air is negligible. We can also solve this question using equations of projectile motion by taking the height of the hill as maximum height, finding theta through it and then applying it in the formula of time of flight.