Question

From the top of a tower of height 10 m, one fire is shot horizontally with a speed of 5*root of 3/s. Another fire is shot upwards at an angle of 60 degrees with the horizontal at some interval of time with the same speed of 5*root of 3 m/s. The shots collide in air at a certain point. The time interval between the 2 shots?

Answer 1

First write equations of motion for both the guns​​​​​​

Gun2: fires at t=0

x direction :
x = 5√3 * cos 60 * (t )

y direction :
y = 5√3 sin60(t) - 1/2 *g *(t)²

Gun1:fires at t = t1

x direction :
x = 5√3 *(t-t1)

y direction :

y = - 1/2 *g *(t-t1)² (negative as upward direction is positive

When they collide their x and y cordinates are equal so if they collide at t=t2 then ,

5√3 *(t2-t1)= 5√3 * cos 60 * (t2 )

- 1/2 *g *(t2-t1)² = 5√3sin60(t2) - 1/2 *g *(t2)²

Solving we get ,

t2 = 2s , t1 =1s

so time interval b/w firing = t1 =1s

Coordinate of point at which both bullet will collide = (5√3 *(2-1) , -1/2*g*(2-1)² )

Point p =(5√3 , -5)