A person can throw a stone to a maximum horizontal distance of 80 m . If the person throws the same stone vertically upwards with the same initial speed, what will be the speed of the stone when it reaches half of its maximum vertical height? (Take g = 10 m/s ^2 )

Step-by-step solution

The maximum horizontal range is obtained when the angle of projection is 45^ . R_ = u^2 g = 80 m u^2 = 80 10 = 800 m ^2 /s ^2 When the stone is thrown vertically upwards, the maximum height attained is: H = u^2 2g = 800 2 10 = 40 m We need to find the speed at half of this maximum height, i.e., at h = 20 m . Using the third equation of motion: v^2 = u^2 - 2gh v^2 = 800 - 2(10)(20) v^2 = 800 - 400 = 400 v = 20 m/s Answer: 20 m/s