(a) 4.2s
(b) 1.73s
Using one of the equations of motion;
h = ut + [tex]\frac{1}{2}[/tex] a[tex]t^{2}[/tex] ------------------(i)
where;
h is the height of the object (rock) relative to the ground,
t = time taken for the object (rock) to travel some distance
a = acceleration due to gravity = g (assume g = 10[tex]m/s^{2}[/tex])
u = initial velocity of the object as it travels down to the ground = 0 (since the rock is dropped there is no initial velocity or the initial velocity is zero(0)).
For the first 88 m fall;
h = 88m
u = 0 (since the rock is being dropped)
a = g = 10[tex]m/s^{2}[/tex] (since the rock drops downwards in the direction of gravity)
Substituting the values of h, u and a into equation (i) we have;
=> 88 = 0(t) + [tex]\frac{1}{2}[/tex] x 10 x [tex]t^{2}[/tex]
=> 88 = 5[tex]t^{2}[/tex]
=> [tex]t^{2}[/tex] = [tex]\frac{88}{5}[/tex]
=> [tex]t^{2}[/tex] = 17.6
Square both sides
=> t = [tex]\sqrt{17.6}[/tex]
=> t = 4.2s
Therefore, time taken for the first 88.0m fall is 4.2s
To do this,
(i) First get the total time taken for the rock to touch the ground from the top of the cliff.
In this case;
h = 176m
u = 0 (since the rock is being dropped)
a = g = 10[tex]m/s^{2}[/tex] (since the rock drops downwards in the direction of gravity)
Substituting the values of h, u and a into equation (i) we have;
=> 176 = 0(t) + [tex]\frac{1}{2}[/tex] x 10 x [tex]t^{2}[/tex]
=> 176 = 5[tex]t^{2}[/tex]
=> [tex]t^{2}[/tex] = [tex]\frac{176}{5}[/tex]
=> [tex]t^{2}[/tex] = 35.2
Square both sides
=> t = [tex]\sqrt{35.2}[/tex]
=> t = 5.93s
Total time (for 176m) is 5.93s
Therefore since;
total time (for 176m) = time for first 88m + time for second 88m
=> time for second 88m = total time (for 176m) - time for first 88m
=> time for second 88m = 5.93 - 4.2
=> time for second 88m = 1.73 seconds
Therefore, time taken for the second 88.0m fall is 1.73s
