Question

The maximum height attained by a ball projected with speed 20

Answer 1

Explanation:

Given,

Given,u = 20m/s

Given,u = 20m/sθ = 45°

Given,u = 20m/sθ = 45°Maximum height in projectile is given by,

Given,u = 20m/sθ = 45°Maximum height in projectile is given by,h = u²sin²θ/2g

Given,u = 20m/sθ = 45°Maximum height in projectile is given by,h = u²sin²θ/2g= (20)²sin²45°/2*9.8

Given,u = 20m/sθ = 45°Maximum height in projectile is given by,h = u²sin²θ/2g= (20)²sin²45°/2*9.8= 100/9.8

Given,u = 20m/sθ = 45°Maximum height in projectile is given by,h = u²sin²θ/2g= (20)²sin²45°/2*9.8= 100/9.8= 10.204 m

Given,u = 20m/sθ = 45°Maximum height in projectile is given by,h = u²sin²θ/2g= (20)²sin²45°/2*9.8= 100/9.8= 10.204 mThus, maximum height attained by the ball is 10.204 m.