Question

a body is thrown from a point with a speed 50 metre per second at an angle 37 degree with horizontal when it has moved the horizontal distance of 80 metre then its distance from point of projection is​

Answer 1

A body is thrown from a point with speed 50m/s at am angle 37° with horizontal.

a/c to question, body moves 80m in horizontal direction. let's find how much time taken to move 80m in horizontal direction.

so, initial velocity of body along horizontal direction is , $u_x=50cos37^{\circ}$

we know, cos37° = 4/5

so, ux = 50 × 4/5 = 40m/s

and acceleration along horizontal direction , ax = 0

now using formula, $x=u_xt+\frac{1}{2}a_xt^2$

or, 80 = 40t => t = 2sec

now find distance travelled during 2 sec in vertical direction.

initial velocity of body along vertical direction, $u_y=50sin37^{\circ}$

= 50 × 3/5 = 30 [as sin37° = 3/5 ]

now using formula, $y=u_yt+\frac{1}{2}a_yt^2$

= 30 × 2 + 1/2 × (-10) × (2)²

= 60 - 20

= 40m

hence, distance travelled by body in vertical direction is 40m