Question

a projectile isto be launched at an angle of 30° so that it falls beyond the pond of length 20 meters as shown in fig
what is the range of values of the initial velocity SO that the projectile falls between point M and N. ​

Answer 1

Answer:

A projectile is to be launched at an angle of 30° so that it falls beyond the pond of length 20 meters as shown in the figure. a) What is the range of values oftheinitial velocity so thattheprojectilefalls betweenpointsMand N?

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I see no diagram. I shall ignore the pond and simply derive launch velocities for range M and range N from the launch point. I must assume M and N are level with the launch point, and that we ignore air resistance.

Let the launch velocity be V at 30 degrees elevation. Then the initial vertical velocity u = V(sin30) = 0.5V. This is continuously reduced by Earth’s gravitational acceleration (9.8 m/s^2), such that instantaneous vertical velocity u = 0.5V -9.8t, where t is time since launch. Re-arrange the equation to get t = (0.5V-u)/9.8. When u becomes 0, the projectile has reached maximum height, and t = (0.5V)/9.8 = 0.051V seconds.

The projectile immediately starts to fall, reversing the trip up: same time, same distance, same changing u. So the total flight time is 2t = 0.102V seconds.

Meanwhile, the initial horizontal velocity x = Vcos30 = 0.866V. There’s no force to change this, so during the entire flight, the projectile moves horizontally at that velocity. Of course, it also moves up then down as described above, to clear stuff on the ground like the lake.

The range (horizontal displacement) of the projectile is that velocity times total flight time. So range R = x(2t) = 0.866V(0.102V) = 0.0883V^2. Then V = 3.365*sqrt(R).

To fall at range M meters, launch velocity V = 3.365sqrt(M) m/s.

To fall at range N meters, launch velocity V = 3.365sqrt(N) m/s.

To fall between M and N, launch velocity can be anything between those numbers.