vf = vi - gt
-2gy= vf 2 – vi 2
y= vi t - (½)gt 2
9. A stone is thrown vertically upward with velocity 40m/s at the edge of a cliff having a height of 110
meters. The stone misses the cliff and continues downward until it hits the ground below.
i) Compute the time it took for the stone to strike the ground below the cliff(=total time in the air).
y(m)
Method 1 )
maximum
First find the time it takes to reach maximium height from
height(v=0)
the cliff(=t1), then find the time it takes for the stone to
fall from the maximum to height to the instant the stone
passes the cliff(=t2), lastly find the time it takes for the
stone to fall the rest of the way down to the ground
below(=t3) and add t1+t2+t3 to get the total time (T).
origin
Step1) Finding t1
vi
vi =40m/s and the when the stone is at the maximum
height vf =0. So from vf = vi – gt, we get
v = -vi
0 = 40 −9.8×t1
Solving t1 we get t1 =4.08s
y=110m
Step2) Finding t2
We do not need to solve t2 since the time it takes for the
stone to come back to its original position is the same as
t1. So t2=4.08s
Step3) Finding t3
The velocity of the stone the moment it is passing the cliff
is −40m/s. We set this velocity as the ‘initial’ velocity and
solve for t3 using y= vi t - (½)gt 2
So inserting vi =−40m/s, y=−110m and g=9.8m/s2 gives
−110=−40t3 - (½)×9.8×t32
Rearranging gives
4.9t32 + 40t3 −110 = 0
Using the quadratic formula (when ax2 + bx +c =0, then x =
t3 =
−𝑏±√𝑏2 −4𝑎𝑐
−40 ±√(−40)2−4×4.9×(−110)
2×4.9
Since time cannot be negative t3=2.17s
Total time T = t1+t2+t3 = 4.08 + 4.08 + 2.17 = 10.3s
1
2𝑎
)
= 2.17s or −10.3s
Method 2) See step3) in Method1. If the initial velocity vi =−40m/s is used to find the time, then the solution
of the quadratic formula will only give the time it takes when the stone is already on its way down passing
the cliff to until it strikes the ground. If that is the case, what if we use vi =+40m/s for the initial velocity?
Since vi=+40m/s represents the speed of the stone going upwards, if we use vi =+40m/s and insert it into
the quadratic formula, then the quadratic formula will give the total time in a single step.
So inserting vi =+40m/s, y=−110m and g=9.8m/s2 gives
−110=40t - (½)×9.8×t2
Rearranging gives
4.9t2 – 40t −110 = 0
−𝑏±√𝑏2 −4𝑎𝑐
Using the quadratic formula (when ax2 + bx +c =0, then x =
t=
40 ±√402 −4×4.9×(−110)
2×4.9
2𝑎
)
= 10.3s or −2.17s
Since time cannot be negative t=10.3s
Method 3) If we find the final velocity vf first, then we can use 'vf = vi - gt' to find the time and avoid using
the quadratic formula.
We can find vf using -2gy= vf 2 – vi 2, so if we rearrange the formula according to vf ,
vf = ±√𝑣𝑖2 − 2𝑔𝑦 =±√(−40)2 − 2 × 9.8 × (−110) = ±61.3m/s
Since the direction of the stone is downwards, we choose the negative. So vf = –61.3m/s
If we rearrange 'vf = vi - gt' according to time,
t=
𝑣𝑓 −𝑣𝑖
−𝑔
=
−61.3−40
−9.8
=10.3s
( Remember that we used vi = +40m/s for the initial velocity. If we use vi = –40m/s and solve ‘t’, then time
will be given as t=2.17s, which is only the time when the stone is on its way down passing the cliff until it
hits the ground)
ii) With what velocity does it strike?
See method3) above
iii) Find the time when the stone is 50m above the cliff(release point) and also find the velocity.
y represents the vertical position from the origin(cliff). So we set y=50m into y= vi t - (½)gt 2and solve for
‘t’ (remember vi = 40m/s and g=9.8m/s2)
50 = 40t – 4.8t2
Solving for t using the quadratic formula gives t=1.54s and 6.62s. Here it is important to understand that
the stone passes 50m two times, on its way up and on its way down. So we keep both times
Answer is t=1.54s(up) and t=6.62s(down).
To find the velocity, we use vf = vi – gt for both times
When t=1.54s, vf = 40– 9.8×1.54 = 25m/s. When t=6.62s, vf = 40– 9.8×6.62 = -25m/s
2
iv) Find the time when the stone is 50m below the cliff and also find the velocity.
We set y= -50m into y= vi t - (½)gt 2and solve for ‘t’ (remember vi = 40m/s and g=9.8m/s2)
-50 = 40t – 4.8t2
Solving for t using the quadratic formula gives t=9.26s and -1.1s.
Time cannot be negative, so only t=9.26s is the answer. And also, we can also see that the stone passes
vertical position y= -50m only once, so there can only be one answer.
Answer is t=9.26s(down).
To find the velocity, we use vf = vi – gt for t=9.26s. So vf = 40– 9.8×9.26 = 51m/s.
3