Question

what must be the horizontal distance of the ball if its thrown with initial velocity of 40m/s at an angle of 20°? pls help

Answer 1

To find:

The horizontal distance of the ball if it is thrown with an initial velocity of 40 m/s at an angle of 20 degrees.

Solution:

It is known that the horizontal distance covered by an object in projectile motion is given by:

$R=\frac{u^2\sin2\theta}{g}$

where u is the initial velocity of the object and theta is the angle of the projectile and g = 9.8 m/s^2.

So, the horizontal distance covered by the object is :

$\begin{gathered} R=\frac{(40)^2sin(40)}{9.8} \\ R=\frac{1600(0.642)}{9.8} \\ R=104.82 \end{gathered}$

Thus, the horizontal distance of the ball is 104.82 meters.