Question

A body is projected at angle 37 degree from horizontal with speed 50 metre per second then find out
1. Time of flight
2. Maximum Height
3. Horizontal Range​

Answer 1

Answer:

at any time t,the projectiles horizontal and vertical displacement are:

x=vo t cos 0 where vo is the initial velocity, 0is the launch angle

y=Vo t sin0 -1/2gt^2

the velocity are

Vx =Vo cos0

Answer 2

$\bold{We \: Know \: That} \\ 1. \: T = \frac{2u \sin \theta }{g} \\ = > T = \frac{2 \times 50 \times \sin(37) }{10} \\ \bold{On \: Putting \: Values \: And \: Cancellation} \\ = > 10 \times \frac{3}{5} \\ = > 6 \: sec \\ 2. \: Hmaximum = \frac{ {u}^{2} { \sin \theta }^{2} }{2g} \\ = > \frac{ {50}^{2} \times ( { \frac{3}{5} })^{2} }{2 \times 10} \\ = > \frac{50 \times 50 \times \frac{3}{5} \times \frac{3}{5} }{20} \\ \bold{On \: Cancellation} \\ = > 5 \times 50 \times \frac{3}{5} \times \frac{3}{5} \times \frac{1}{2} \\ = > 45m \\ 3. \: R = \frac{ {u}^{2} \times \sin2 \theta }{g} \\ = > \frac{ {50}^{2} \times 2 \times \sin37 \times \sin37 }{10} \\ = > 50 \times 50 \times \frac{4}{5} \times \frac{4}{5} \times \frac{1}{10} \times 2 \\ = > R = 240m$