Question

A projectile starting from ground hits a target on the ground located

at a distance of 1000 meters after 40 seconds.

a) What is the size of the angle θ?

b) At what initial velocity was the projectile launched?​

Answer 1

Answer:

Given

horizontal range (r) = 1000 m

time of flight (t) = 40 s

we know that

$r = u \cos( \theta ) t \\ 1000 = u \cos( \theta ) 40 \\ u \cos( \theta ) = 25m {s}^{ - 1} \: \: \: \: \: \: \: eq \: 1 \\ \\ t = \frac{2u \sin( \theta ) }{g} \\ 40 = \frac{2u \sin( \theta ) }{10} \\ u \sin( \theta ) = 200m {s}^{ - 1} \: \: \: \: \: \: \: eq \: 2 \\ \\ eq \: 2 \div eq \: 1 \\ \frac{u \sin( \theta ) }{u \cos( \theta ) } = \frac{200}{25} \\ \tan( \theta ) = 8 \\ \theta = {tan}^{ - 1} (8) \\ \\ \cos( \theta ) = \frac{1}{ \sqrt{65} } \\ from \: eq \: 1 \\ u \cos( \theta ) = 25 \\ u = 25 \sqrt{65} m {s}^{ - 1}$