ENGG – DYNAMICS OF RIGID BODIES
1. The jet plane travels along the vertical parabolic path. When it is at point A it has a
speed of 200 m/s, which is increasing at the rate of 0.8 m/s2. Determine the
magnitude of acceleration of the plane when it is at point A.
GIVEN:
v = 200 m/s
𝑎𝑡 = 0.8 m/𝑠
y = 0.4𝑥
2
2
REQUIRED:
Magnitude of acceleration = ?
EQUATION/S:
𝑑𝑦 2 3/2
ρ=
[1+( 𝑑𝑥 ) ]
2
𝑑𝑦
2
𝑑𝑥
2
𝑎𝑛 =
𝑉
ρ
SOLUTION:
(since it can be seen from the illustration that it forms a parabola)
𝑑𝑦
𝑑𝑥
For
:
Find the derivative of y = 0.4𝑥
𝑑𝑦
𝑑𝑥
𝑑𝑦
𝑑𝑥
2
= 2(0. 4𝑥)
= 0. 8𝑥
Find the second derivative of
𝑑𝑦
𝑑𝑥
= 0. 8𝑥
2
𝑑𝑦
2
𝑑𝑥
= 0. 8
𝑑𝑦 2 3/2
Substitute values to ρ =
[1+( 𝑑𝑥 ) ]
2
𝑑𝑦
:
2
𝑑𝑥
2 3/2
ρ=
[1+(0.8𝑥) ]
3
= 87. 62 𝑘𝑚 , where x = 5km or 5 × 10 m (based on the
0.8
illustration given)
2
𝑉
ρ
Solve for 𝑎𝑛 =
:
2
𝑎𝑛 =
(200𝑚/𝑠)
= 0. 456 𝑚/𝑠
3
87.62×10 𝑚
2
Solve for the acceleration of the plane at Point A:
2
2
𝑎=
𝑎𝑡 + 𝑎𝑛
𝑎=
(0. 8𝑚/𝑠 ) + (0. 456 𝑚/𝑠 )
2 2
𝑎 = 0. 921 𝑚/𝑠
2
2
2
ANSWER:
𝑎 = 0. 92 𝑚/𝑠
2
2. Speedy drives horizontally off a cliff at 15 m/s. He lands 30m from the base of the
cliff. How high is the cliff?
GIVEN:
v = 15 m/s (speed)
d = 30m (from the base)
REQUIRED:
Height of the cliff = ?
EQUATION/S:
t = distance/velocity or speed
𝑦 = 𝑦0 + 𝑣0𝑡 +
1
2
2
𝑎𝑡 , where
1
2
≈ 0. 5
1
2
≈ 0. 5
SOLUTION:
Compute for time:
𝑡=
𝑡=
𝑡=
𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒
𝑠𝑝𝑒𝑒𝑑
30 𝑚
2
15 𝑚/𝑠
30 𝑚
15 𝑚/𝑠
𝑡 = 2𝑠
Compute for the height:
𝑦 = 𝑦0 + 𝑣0𝑡 +
𝑦 = 0. 5(− 9. 81
1
2
2
𝑎𝑡 , where
𝑚
2
𝑠
2
)(2𝑠)
𝑦 = 19. 62𝑚
ANSWER:
The cliff is 19. 62𝑚 high.
3. On the planet Mars, the gravitational acceleration is 3.7 m/s2. Suppose a rock is
thrown upward at a speed of 20 m/s at an angle of 70 degrees above the horizontal.
Find the time of the rock's flight and the maximum height it reaches.
GIVEN:
g = 3. 7 𝑚/𝑠
v = 20 m/s
2
◦
θ = 70 𝑎𝑏𝑜𝑣𝑒 𝑡ℎ𝑒 ℎ𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙
REQUIRED:
𝑡= ?
ℎ= ?
EQUATION/S:
𝑇𝑖𝑚𝑒 𝑜𝑓 𝐹𝑙𝑖𝑔ℎ𝑡 (𝑡) =
2𝑣𝑠𝑖𝑛θ
𝑔
2
𝑀𝑎𝑥𝑖𝑚𝑢𝑚 𝐻𝑒𝑖𝑔ℎ𝑡 (ℎ𝑚𝑎𝑥) =
2
𝑣 𝑠𝑖𝑛 θ
2𝑔
SOLUTION:
Solve for the time of flight:
𝑇𝑖𝑚𝑒 𝑜𝑓 𝐹𝑙𝑖𝑔ℎ𝑡 (𝑡) =
2𝑣𝑠𝑖𝑛θ
𝑔
𝑇𝑖𝑚𝑒 𝑜𝑓 𝐹𝑙𝑖𝑔ℎ𝑡 (𝑡) =
2(20 𝑚/𝑠) 𝑠𝑖𝑛(70 )
◦
2
3.7 𝑚/𝑠
𝑡 = 10. 159 𝑠
Solve for the maximum height:
2
2
ℎ𝑚𝑎𝑥 =
𝑣 𝑠𝑖𝑛 θ
2𝑔
ℎ𝑚𝑎𝑥 =
(20 𝑚/𝑠) 𝑠𝑖𝑛 70
2
2
◦
2
2(3.7 𝑚/𝑠 )
ℎ𝑚𝑎𝑥 = 47. 731𝑚
ANSWER:
𝑡 = 10. 16 𝑠
ℎ𝑚𝑎𝑥 = 47. 73𝑚
4. A catapult launches a rock at a castle wall. The rock is launched at a 45-degree angle
with a speed of 22 m/s. The castle wall is 45m away and is 20m high. Does the rock
make it over the wall?
GIVEN:
◦
θ = 45
v = 22 m/s
d = x = 45m (away, so at point x)
h = 20m (hight, so at point y)
EQUATION/S:
𝑣𝑜𝑥 = 𝑣𝑐𝑜𝑠θ
𝑣𝑜𝑦 = 𝑣𝑠𝑖𝑛θ
𝑡=
𝑥
𝑣
=
𝑦 = 𝑦0 +
𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒
𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑜𝑟 𝑠𝑝𝑒𝑒𝑑
2
1
𝑣0𝑡 + 2 𝑎𝑡
SOLUTION:
◦
𝑣𝑜𝑥 = (22𝑚/𝑠)𝑐𝑜𝑠(45 ) = 11 2 ≈ 15. 556 𝑚/𝑠
◦
𝑣𝑜𝑦 = (22𝑚/𝑠)𝑠𝑖𝑛(45 ) = 11 2 ≈ 15. 556 𝑚/𝑠
@ the horizontal
◦
𝑣𝑜𝑥 = (22𝑚/𝑠)𝑐𝑜𝑠(45 ) = 11 2 ≈ 15. 556 𝑚/𝑠
𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡 𝑎𝑡 𝑥0 = 0
𝑥 = 45𝑚 (𝑎𝑤𝑎𝑦)
Substituting the values for horizontal;
𝑡=
𝑡=
𝑡=
𝑡=
𝑥
𝑣
𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒
𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑜𝑟 𝑠𝑝𝑒𝑒𝑑
=
45𝑚
11 2 𝑚/𝑠
45𝑚
11 2 𝑚/𝑠
45 2
22
𝑠 ≈ 2. 893 𝑠
@ the vertical
◦
𝑣𝑜𝑦 = (22𝑚/𝑠)𝑠𝑖𝑛(45 ) = 11 2 ≈ 15. 556 𝑚/𝑠
2
𝑎 =− 9. 81 𝑚/𝑠 (𝑠𝑖𝑛𝑐𝑒 𝑖𝑡'𝑠 𝑎𝑔𝑎𝑖𝑛𝑠𝑡 𝑡ℎ𝑒 𝑎𝑖𝑟)
𝑦𝑜 = 0
𝑡 = 2. 893 𝑠
Substitute the values to the equation:
𝑦 = 𝑦0 + 𝑣0𝑡 +
1
2
2
𝑎𝑡
𝑦 = 0 + (15. 556 𝑚/𝑠)(2. 893 𝑠) +
1
2
2
(− 9. 81 𝑚/𝑠 )(2. 893 𝑠)
2
𝑦 = 3. 95 𝑚
ANSWER:
No, the rock did not make it over the wall since the wall’s height is 20 meters high,
but when we got the value of y, we just obtained a sum of 3.95 m.