Question

A ball is projected from ground level with a speed of 28 ms1 at an angle of 30° to the horizontal as is a wall of height h at a distance of 16 m from the point of projection of the resistance is negligible.

Answer 1

To find the height of the wall (h), we can use the equations of projectile motion.

First, we need to break down the initial velocity of the ball into its horizontal and vertical components.

The horizontal component (Vx) can be found using the equation Vx = V * cos(theta), where V is the initial speed (28 m/s) and theta is the angle of projection (30°).
Vx = 28 m/s * cos(30°)
Vx = 24.248 m/s

The vertical component (Vy) can be found using the equation Vy = V * sin(theta).
Vy = 28 m/s * sin(30°)
Vy = 14 m/s

Since the resistance is negligible, the vertical motion of the ball is affected by gravity only.

Using the equation for vertical displacement, we have:
h = (Vy^2) / (2 * g)
where g is the acceleration due to gravity (approximately 9.8 m/s^2).
h = (14 m/s)^2 / (2 * 9.8 m/s^2)
h = 196 m^2/s^2 / 19.6 m/s^2
h = 10 m

Therefore, the height of the wall (h) is approximately 10 meters.