Question

a ball is thrown from a field with a speed of 12 m/s at an angle of 45° with the horizontal. at what distance will it hit the field again ?

Answer 1

The formula for RANGE (of a projectile) is given by: V^2 sin 2 (theta) / g,

Since theta is 45, (2) theta = 90, sin 90 = 1.

Range = 12^2/g = 144/10 = 14.4 m

Answer 2

HELLO THERE!

The given question belongs to "Projectile Motion".

Initial velocity with which the ball is thrown (u) = 12 ms⁻¹

Angle with the horizontal (θ) = 45°

The question asks the Range of the Projectile Motion, i.e., the horizontal distance covered by the ball.

The formula to calculate the range is:

$R = \frac{u^{2}sin2\theta}{g}\\\\\\= \frac{(12)^{2} \times sin(2\times45)}{10}\\\\\\= \frac{144}{10}\\\\= 14.4m$

So, the ball will hit the ground after travelling 14.4 m.

THANKS!