Answer:
[tex]45m[/tex]
Explanation:
Let [tex]V_0[/tex] be the horizontal speed of the keys and D be the horizontal distance covered by the keys.
[tex]V_0= 8.0 \frac ms[/tex]
[tex]D= 28 m[/tex]
The initial velocity in the vertical direction is [tex]0[/tex].
Let [tex]g[/tex] be the acceleration due to gravity, which always acts vertically downwards, so, it will not change the horizontal direction of the speed, i.e. [tex]V_0[/tex] will remain constant throughout the projectile motion.
Let [tex]t[/tex] be the total time of flight in seconds. So,
[tex]D= V_0t[/tex]
[tex]\Rightarrow 28=8t[/tex]
[tex]\Rightarrow t=3 s \; \cdots (i)[/tex]
Let [tex]H[/tex] be the hight of the cliff.
Now, from the equation of motion
[tex]s=ut+\frac 1 2 at^2\;\cdots(ii)[/tex]
Where is the displacement is the direction of force, is the initial velocity in the direction of force, is the constant acceleration due to force and is the time of flight.
Here, , and (negative sign is for taking the sigh convention positive in the direction as shown in the figure.)
From equation (ii)
[tex]-H=0\times t + \frac 1 2 (-g)t^2[/tex]
[tex]\Rightarrow H= \frac 1 2 gt^2[/tex]
Take [tex]g= 10 \frac m{s^2}[/tex] and put the value of [tex]t[/tex] from equation (i), we have
[tex]H=\frac 1 2 \times 10\times 3^2[/tex]
[tex]\Rightarrow H= 45 m[/tex].
Hence, the height of the cliff was [tex]45 m[/tex].
