A ball is projected horizontally from the top of a hill with a velocity of 30ms-1. If it reaches the ground 5 sec later. The height of the hill is what?

LawsonOche

6 years ago

s = u t + ½ a t². In this expression u= 0 m/s, a = 10m/
s², t = 5 s. Substituting in the equation we get, s = 0× t + ½ ×10 m/s²× 5²s²= 5× 25 m = 125 m. The height of the hill is 125m.

Sydneystudio

6 years ago

Assuming there is no air resistance, the
horizontal component of the velocity has
no effect on the vertical component or the
time of flight (on horizontal ground).
But this is on a hill. Does the ball reach the
horizontal plane below the hill, or land on
the hillside, or does it land on the
neighbouring hill? Let’s change the hill to
a tower on a perfectly horizontal plane.
In that case this is the same as the ball
falling from the given height. Use the
equation . The initial
velocity in this case is the vertical
component and is either , or
depending on your sign

ojehlc

2 years ago

H= UT
Therefore, 30*5= 150m