A shot is fired with a bullet with an initial velocity of 20 m/s from a point 10 m infront of a vertical wall 5 m high.

Find the angle of projection with the horizontal to enable the shot to just clear the wall. Also find the range of the shot where the bullet falls on the ground.

**Given** : u=20 m/s

Distance from wall=10m

Height of wall=5m**To find** :Angle of projection

Range of shot

#### Solution

Let α be the angle of projection of projectile

Equation of projectile is given by:

y=**xtan**α - `(gx^2)/(2u^2) sec^2α`

(10,5) are the co-ordinates of top of wall when O is taken as origin

Substituting x=10 and y=5 in the projectile equation 5=10tanα-`(g10^2)/(2 xx 20^2)sec^2α`

1.2262tan2 α-10tan α+6.2262=0

Solving the quadratic equation

**tan α=7.4758 or tan α=0.6792**

α=82.381°or α=34.184°

Range of a projectile is given by R = `(u^2sin2alpha)/g`

Substituting α=82.381° or α=34.184°

`R=(20^2sin(2xx82.381))/ 9.81`

=10.7161 m

`R=(20^2sin(2 xx 34.184)0)/9.81`

=37.902 m

Angle of projectile should be 82.381° or 34.184° and the corresponding ranges will be 10.7161 m and 37.902 m respectively.