91Ó°ÊÓ

Here angle is$\mathrm{θ}=45°$and initial is$v_0$ .

This problem is based on the concept of projectile motion. When a particle is thrown near the earth’s surface, it travels along a curved path under constant acceleration directed towards the center of the earth surface. Using concept of projectile, the range of projectile can be found. For range, vertical and horizontal components of initial velocity as well as vertical and horizontal distances can be used. Also using the kinematic equations the time of flight of projectile can be found.

Formulae:

$R=\frac{v_0^2×\sin \left(2\mathrm{θ}\right)}{g}$

$t=\frac{2v_{0}s\mathrm{in}\left(\mathrm{θ}\right)}{g}$

At$\mathrm{θ}=45°$ every projectile has maximum range

We can find the range R of the projectile by plugging the values of initial velocity and angle.

As the angle increased the range will decrease because of$\sin \left(2\mathrm{θ}\right)$ . As angle increases, the value of sin$\left(2\mathrm{θ}\right)$ will decrease.

If the angle is increased the value of t increases as the higher angles value of sin tends to 1.