Question

A particle is projected from the ground with some initial velocity making an angle of 45° with the horizontal. It reaches a height of 7.5 m above the ground while it travels a horizontal distance of 10 m from the point of projection. Find the initial speed of projection. (g = 10 m/s²)

Answer 1

it is based on the concepts of oblique projectile motion.

use formula, $\bf{y=xtan\theta-\frac{gx^2}{2u^2cos^2\theta}}$

here, y = 7.5m , x = 10m , g = 10m/s²
$\theta=45^{\circ}$

7.5 = 10tan45° - 10 × 10²/(2 × u² × cos²45°)

7.5 = 10 - 1000/{2 × u² × (1/√2)²}

-2.5 = -1000/u²

2.5 = 1000/u²

u² = 400 = (20)²

u = 20m/s

hence, initial speed of projection 20m/s

Answer 2

Hii dear,

# Answer- 1 m/s

## Explaination-
# Given-
R = 10 m
H = 7.5 m
θ = 45°

# Formula-
Range of projectile is given by,
R = u^2sin2θ / g
Putting values,
10 = u^2sin90 / 10
u^2 = 1
u = 1 m/s

Initial velocity of projection of particle is 1 m/s.

Hope that is useful.