A particle is projected horizontally at 15ms(^{-1}) from a height of 20m. Calculate the horizontal distance covered by the particle just before hitting the ground. [g= 10ms(^{-2})]
Explanation
Let R represent the horizontal distance covered at time t.
Using h = ut + (frac{1}{2}gt^2)
but U(_v) = 0
h = (frac{1}{2} gt^{2})
20 = (frac{1}{2} times 10 times t^2)
20 = 5t(^2)
t(^2) = (frac{20}{5}) = 4
t = 2s
Range(R) = U(_x) x t
= 15 x 2
= 30m
or
R = u x (sqrt{frac{2h}{g}})
= 15 x (sqrt{frac{2 times 20}{10}})
= 15 x (sqrt{4})
= 15 x 2
= 30m