Question

1. A warship has a gun that fires a projectile at an initial speed of 400 m/s. It fires at an enemy ship that is 10.0 km away. At what angle from the horizontal should the gun be fired in order to hit the enemy, assuming that the enemy’s location does not change? (Hint: you can write two equations involving time, one for the horizontal motion, and one for the vertical motion. Solve one equation for time and substitute into the other, in order to eliminate time & then solve for the angle. You can use the trigonometric identity sin 2? = 2 sin ? cos ?.)

2. An airplane is attempting to land on a runway when the wind is blowing at a velocity of 10 m/s perpendicular to the runway. Given that the airplane is flying at an airspeed of 47 m/s, at what angle relative to the runway direction must the pilot keep the nose pointed into the wind to maintain a flight path aligned with the runway?

3. You park your buddy’s car on an incline that makes an angle of 17.0° below the horizontal, sloping toward a cliff of height 55.0 m, overlooking the ocean, but you forget to set the brake. While you are walking and enjoying the view, the car’s transmission slips out of park, and it begins rolling toward the cliff edge. Assume zero rolling friction and zero air resistance. Given that the car rolls 29.0 m before flying off the cliff,

   1. What is the car’s velocity just as it leaves solid ground?

   2. For how long is the car in the air?

   3. How far away from the base of the cliff does it strike the water?

Answer 1

1. If u be the initial velocity, R be the range, T be the time of flight, then, u cos is the horizontal component and u sin is the vertical component of initial velocity. Considering horizontal motion from initial to final point, we get, R = u cos*T .................(1)

and for vertical motion from initial to the highest point attained during flight (where final velocity, v = 0), we have,

v = u sin - gt => 0 = u sin - gT/2 => 2u sin = gT..... (2)

Now, (1)/(2) => R/2u sin = u cos*T/gT => Rg/u² = sin 2

Here, we use, 2 sin cos = sin 2, thus, 2 = sin^-1( Rg/u² )

Or, 2 = sin^-1( 10*10³*9.8/400² ) = 37.78⁰

Hence, initial angle of launch, = 37.78⁰/2 = 18.89⁰