Question

. A rocket was launched to hit the terrorists’ hideout at a velocity of 15,000 ft/s at a

target 300 miles away.

a.) At what angle must it be fired to hit the target?

b.) How long after it is fired will the target be hit?

c.) If the terrorists discovered the plot to eliminate them 2 minutes after the rocket

was launched, will they escape alive? ​

Answer 1

Answer:

Given

initial velocity of rocket(u) = 15,000 ft/s

(1 ft = 0.305 m) = 4,575 m/s

target (horizontal range(r)) = 300 miles

= 480 km

a) angle of projection

$r = \frac{ {u}^{2} \sin( 2\theta ) }{g} \\ 480000 = \frac{4575 \times 4575 \times \sin(2 \theta) }{10} \\ \sin(2 \theta ) = 0.2 \\ \sin(2 \theta ) = \frac{1}{5} \\ \theta = \frac{1}{2} { \sin}^{ - 1} ( \frac{1}{5} )$

b) time of flight

$\sin( \theta ) = 0.1(approx) \\ t = \frac{2u \sin( \theta ) }{g} \\ t = \frac{2 \times 4575 \times 0.1}{10} \\ t = 91.5s$

c) the rocket launched will hit the target in 1.525 minutes.

So, they can't escape alive