Answer:
The initial speed of the ball is 30 m/s.
Explanation:
It can be assumed that the ball is thrown at an angle of 45 degrees to the ground. The ball lands 90 m away. We need to find the initial speed of the ball. We know that the horizontal distance covered by the projectile is called its range. It is given by :
[tex]R=\dfrac{u^2\ sin2\theta}{g}[/tex]
u is the initial speed of the ball.
[tex]v=\sqrt{\dfrac{Rg}{sin2\theta}}[/tex]
[tex]v=\sqrt{\dfrac{90\times 9.8}{sin2(45)}}[/tex]
v = 29.69 m/s
or
v = 30 m/s
So, the initial speed of the ball is 30 m/s. Hence, this is the required solution.
