Monday, January 19, 2009

Problems- Bodies in vertical motion.

1. A stone is projected vertically upwards with a velocity 29.4m/sec. i)Calculate the maximum height to which it rises ii) calculate the time taken to reach maximum height iii) Find the ratio of velocities after 1sec, 2 sec, 3sec of its journey.

Soln: From the data given in the problem,

Initial velocity of stone u=29.4 m/sec,

Acceleration due to gravity g=9.8 m/sec^2,

i) H_{max} = \frac{u^2}{2g} = \frac{(29.4)^2}{2(9.8)}

H_{max} = \frac{29.4}{2} = 14.7 m.

ii) Time of ascent T_a =\frac{u}{g} = \frac{29.4}{9.8} = 3sec.

iii) Ratio of velocities of a vertically projected up body after 1 sec, 2 sec, 3sec , . . . . . . . . . . . . . . of its journey will be v_1:v_2:v_3 . . . . . . . . . . . . :v_n = (u-g) : (u-2g) : (u-3g) : . . . . . . . . : (u-ng).

v_1:v_2:v_3 = (29.4-9.8 ) : (29.4- 19.6) : (29.4 - 29.4)

v_1:v_2:v_3= 19.6 : 9.8 : 0 .

2. A stone is dropped from the top of a tower.The stone touches the ground after 5sec.Calculate the i)height of the tower and the ii) velocity with which it strikes the ground.

Soln: From the data given in the problem,

Initial velocity of the stone u = 0 m/sec,

Acceleration due to gravity g = 9.8 m/sec^2

Time of descent t_a = 5 sec.

i) Height of the tower be S=H (say)

substitute the above values in the equation S=ut+ \frac{1}{2}at^2

H = 0(5)+\frac{1}{2}(9.8)(5)^2 = 0 + \frac{1}{2}(9.8)(25) = (4.9)(25) = 122.5 m.

ii) Let the velocity when it touches = v (say),

substitute the above values in the equation v=u +gt_a

v = 0+(9.8)(5) = 49 m/sec.

3. A stone is projected vertically upwards with an initial velocity 98 m/sec . Calculate i) maximum height the body reaches ii ) find its velocity when it is exactly at the mid point of it’s journey iii) time of flight.

Soln : From the data given in the problem,

Initial velocity of the stone u = 98 m/sec,

acceleration due to gravity a = 9.8 m/sec^2,

i) Maximum height H_{max} = \frac{u^2}{2g} = \frac{(98)^2}{2(9.8)} = 490 m.

ii) When it is in the middle S= H_{max}{2} =245 m

Let the velocity = v (say),

substitute the values in the equation v^2 - u^2 = 2gS,

we get v^2 - (98)^2 = 2(-9.8)(245),

v^2 = 9604 - 4802 =4802 ,

v =69.30 m/sec

iii) time of flight T = \frac{2u}{g} = \frac{2(98)}{(9.8)} = 20 sec.

4.Estimate the following.
(a) how long it took King Kong to fall straight down from the top of a 360 m high building?_________ seconds
(b) his velocity just before “landing”___________ m/s. ( Question by ELISE in Yahoo answers).

Soln: From the data given in the problem,

Height of building S=H = 360m,

Acceleration Due to gravity g=10 m/sec^2,

i) time of descent be t_d= ?

substitute the values in the formula S=ut+\frac{1}{2}gt^2

we get 360 = (0)(t_d) +\frac{1}{2}(10)(t_d)^2 ,

360 = 5 t_d^2 ; t_d^2 = 360/5=72

Therefore t_d = 8.49 sec.

ii )Velocity of king kong just before touching the ground v =gt_d,

v = (10)(8.49) =84.9 m/sec.

5. A baseball is hit straight up into the air with a speed of 23 m/s.
(a) How high does it go?____ m
(b) How long is it in the air?_____ s. ( Question by ELISE in Yahoo answers).

Soln: From the data given in the problem,

Acceleration Due to gravity g=10 m/sec^2,

a) Maximum height H_{max}= \frac{u^2}{2g} = \frac{(23)^2}{20}

H_{max}= 529/20 = 26.45 m.

b ) Time of flight of base ball T = \frac{2u}{g},

T= \frac{(2)(23)}{10}, = 46/10 = 4.6 sec.


Posted by chavan gyanraj at 10:56 PM
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