Thursday, January 29, 2009

Problems-Bodies in vertical motion

6.A body starting from rest slides down an inclined plane.Find the velocity after it has descended vertically a distance of 5metres. (g=9.8 m/sec^2).

Soln: The velocity of the body when it touches the ground sliding down an inclined plane ,will be same as when the body vertically falls freely from height ‘h”.

From the problem height S=h=5m; acceleration due to gravity g=9.8m/sec^2; Initial velocity u=0 ; V=?

substitute the values in the equation V^2-u^2 = 2gs ,

we get V^2-0^2 = 2(9.80) (5) ; V^2 = 98

V =9.899 m/sec.

7.A ball projected vertically upwards returns to ground after 15sec.Calculate i)maximum height to which it rises ii)velocity with which it is projected and iii)its position after 6 seconds.

Soln:From the data given in the problem,

Velocity of projection u=?

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

time of flight T =15 sec

substitute the values of T,g in the equation T = \frac{2u}{g}

\frac{2u}{9.8} =15 ; 2u = 15 (9.80) = 147

ii )Velocity with which it is projected u= 73.5 m/sec.

i)Maximum height H_{max} =\frac{u^2}{2g},

substitute the values of u,g in the equation we get H_{max} = \frac{73.5^2}{2(9.8)}

H_{max} = \frac{5402.25}{19.6} =275.625 m

1iii) Let Its position after t=6sec be S

substitute the values of u,t and g in the equation S=ut - \frac{1}{2} at^2

S = (73.5)(6) - \frac{1}{2} 9.8(6)^2=441-176.4 = 264.6 m high from the ground.

8.A stone was dropped from a rising baloon at a height of 150m above the ground and it reaches the ground in 15 seconds.Find the velocity of the baloon at the instant the stone was dropped.

Soln: From the data given in the problem,

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

Height of the balloon when the stone is dropped from it h=150m,

Time of flight T=15 sec,

Let the velocity of the baloon when the stone is dropped from it is u (say).

Substitute the values of g,t and h in the equation h = \frac{1}{2} gt^2 -ut

we get 150 = \frac{1}{2} (9.8)(15)^2-u(15) = (4.9)(225) - 15u,

150 = 15[(4.9)(15)- u],

10=73.5 - u ; u = 73.5- 10 =63.5 m/sec.

9. From the top of a tower of 200 metres high, a stone is projected vertically upwards with a velocity 49m/sec.Calculate the i)maximum height traveled by it from ground level, ii)the velocity with which it strikes the ground and the iii) time it takes to reach the ground.

Soln: From the data given in the problem,Height of the tower is h=200m,


velocity of projection u = 49 m/sec,

Max height of the stone from top of the tower be S= X (say),

velocity at maximum height v = 0,

i) substitute the values of u,v and g in equation V^2-u^2 = -2gs,

we get 0^2-49^2 = -2(9.8)X,

X = \frac{2401}{19.6} = 122.5 m from the top of the tower

Maximum height from ground level H = X +h =122.5 + 200 = 322.5 m.

ii) Let the velocity with which it reaches the ground be v_1 say,

substitute the values in the equation v_1= \sqrt{2gH}

v_1= \sqrt{2(9.8)(322.5)},

v_1= \sqrt{6321},

v_1= 79.50 m/sec.

iii) Time of flight T =?

Substitute the values u,g and h in the equation h = \frac{1}{2} gt^2 -ut,

we get 200 = \frac{1}{2} (9.8)T^2 -49T,

200 = (4.9)T^2 - 49T

T^2- 10T = \frac{200}{4.9},

by solving this quadratic equation for T we get T= 13.11 sec

Posted by chavan gyanraj at 2:34 AM
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1 comment:

  1. AnonymousDecember 30, 2009 at 11:32 PM

    Vertical motion under gravity is a specific case of one dimensional motion with constant acceleration. Here, acceleration is always directed in vertically downward direction and its magnitude is "g". As the force due to gravity may be opposite to the direction of motion, there exists the possibility that the body under force of gravity reverses its direction. It is, therefore, important to understand that the quantities involved in the equations of motion may evaluate to positive or negative values with the exception of time (t). We must appropriately assign sign to various inputs that goes into the equation and correctly interpret the result with reference to the assumed positive direction. Further, some of them evaluate to two values one for one direction and another of reversed direction. As pointed out earlier in the course, we must also realize that a change in reference direction may actually change the sign of the attributes, but their physical interpretation remains same. What it means that an attribute such as velocity, for example, can be either 5 m/s or -5 m/s, conveying the same velocity. The interpretation must be done with respect to the assigned positive reference direction. I am a college sophomore with a dual major in Physics and Mathematics @ University of California, Santa Barbara. By the way, i came across these excellent physics flash cards. Its also a great initiative by the FunnelBrain team. Amazing!!

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