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

Answer:

a) Time = 2.67 s

b) Height = 35.0 m

Explanation:

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

$x_{f} = x_{0} + v_{0_{x}}t + \frac{1}{2}at^{2}$   (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:  

$t = \frac{x_{f}}{v_{0}} = \frac{80 m}{30 m/s} = 2.67 s$

b) The height of the hill is given by:

$y_{f} = y_{0} + v_{0_{y}}t - \frac{1}{2}gt^{2}$

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:

$y_{0} = \frac{1}{2}gt^{2} = \frac{1}{2}9.81 m/s^{2}*(2.67 s)^{2} = 35.0 m$  

I hope it helps you!