Astone is proyected vertically upward with a unknown spo ed from a poat onto pofatal tee ID: A having a beight of H-15 m, as shown below. The stone reaches the groand afher b 8 s. is the initial velocity of the stone? Uise the labels given to solve the problem. ?.vt-lt 6m Vi 31.325 b) How long does it take the stone to reach point 4 which is 6 m above the ground? Ay- 15 39,7 32.1 31316-4.3t
Solution
(a) The distance covered by the stone = - 15 m
-ve sign indicates the direction that is opposite to the initial motion.
Let us consider the initial velocity of the stone = u
time taken by the distance (t ) = 8 s
Therefore using the equation of motion
S = ut + (1/2)at 2
where S is displacement , a is acceleration -9.81 m/s 2
-15 = u*8 + 1/2(-9.81)*8 2
-15 = 8u - (1/2)*9.81*64
u = 37.365 m/s
(b) When the stone reaches at the point 4 i.e 6 m above the ground , therefore displacement will be
S = 6 -15 = -9 m
Now applying the equation of motion
S = ut + (1/2)at 2
-9 = 37.365*t + (1/2)*(-9.81)*t 2
-9 = 37.365t - 4.9t 2
4.9t 2 - 37.365t - 9 = 0
on solving we get
t = 7.86 s
.
Influencing policy (training slides from Fast Track Impact)
Astone is proyected vertically upward with a unknown spo ed from a poa.docx
1. Astone is proyected vertically upward with a unknown spo ed from a poat onto pofatal tee ID: A
having a beight of H-15 m, as shown below. The stone reaches the groand afher b 8 s. is the
initial velocity of the stone? Uise the labels given to solve the problem. ?.vt-lt 6m Vi 31.325 b)
How long does it take the stone to reach point 4 which is 6 m above the ground? Ay- 15 39,7
32.1 31316-4.3t
Solution
(a) The distance covered by the stone = - 15 m
-ve sign indicates the direction that is opposite to the initial motion.
Let us consider the initial velocity of the stone = u
time taken by the distance (t ) = 8 s
Therefore using the equation of motion
S = ut + (1/2)at 2
where S is displacement , a is acceleration -9.81 m/s 2
-15 = u*8 + 1/2(-9.81)*8 2
-15 = 8u - (1/2)*9.81*64
u = 37.365 m/s
(b) When the stone reaches at the point 4 i.e 6 m above the ground , therefore displacement will
be
S = 6 -15 = -9 m
Now applying the equation of motion
S = ut + (1/2)at 2
-9 = 37.365*t + (1/2)*(-9.81)*t 2
-9 = 37.365t - 4.9t 2
4.9t 2
- 37.365t - 9 = 0
on solving we get
t = 7.86 s
