Question

a stone is thrown horizonttaly from a cliff of a hill with an initial velocity of 30m/s it hits the ground at a horizontal distance of 80m from the foot of the hill how long does a stone travel? what is the height of the hill? (please show the process)​

Answer 1

a) Time = 2.67 s

b) Height = 35.0 m

Step-by-step explanation:

a) The time of flight can be found using the following equation:

`x_(f) = x_(0) + v_{0_(x)}t + (1)/(2)at^(2)` (1)

Where:

`x_(f)`: is the final position in the horizontal direction = 80 m

`x_(0)`: is the initial position in the horizontal direction = 0

`v_{0_(x)}`: is the initial velocity in the horizontal direction = 30 m/s

a: is the acceleration in the horizontal direction = 0 (the stone is only accelerated by gravity)

t: is the time =?

By entering the above values into equation (1) and solving for "t", we can find the time of flight of the stone:

`t = (x_(f))/(v_(0)) = (80 m)/(30 m/s) = 2.67 s`

b) The height of the hill is given by:

`y_(f) = y_(0) + v_{0_(y)}t - (1)/(2)gt^(2)`

Where:

`y_(f)`: is the final position in the vertical direction = 0

`y_(0)`: is the initial position in the vertical direction =?

`v_{0_(y)}`: is the initial velocity in the vertical direction =0 (the stone is thrown horizontally)

g: is the acceleration due to gravity = 9.81 m/s²

Hence, the height of the hill is:

`y_(0) = (1)/(2)gt^(2) = (1)/(2)9.81 m/s^(2)*(2.67 s)^(2) = 35.0 m`

I hope it helps you!