CALCULATING OBJECTS THROWN IN THE AIR !link to the graph: http://www.mediafire.com/imageview.php?quickkey=sda35fx1cibzxgl&undefined&r=ustag 1) Discuss the significant features of the...

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neela | High School Teacher | (Level 3) Valedictorian

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a)

Definitely there is a pattern in the graph. A stone projected upwards , reaches greatest height  125 (appr)  at around 5 seconds and then begins to fall. If we draw a vertical from the vertex (greatest height of 125 around) ,  then the curve is symmetrical about this vertical line. The the height travelled in every succesive equal interval is diminishing till the greatest height is reached and then it increases while falling.

The  values  taken below is from the graph by the eye judgement and so is approximate only.

b)

At t=2 and t = 8, the graph is very clearly near round figure.

For us it looks that at t = 2,  H(2) = 79 and at t = 8 , H(8) = 81.

Also we know that H(t) = ut-(1/2)gt^2 , where u is the initial velocity , g is the acceleration due to gravitation.

We estimate the u and g from the graphical positions  of (t, H(t)) at  t = 2 and t =8 which looks better accurate points from visual accuracy.

H(t) = ut-(1/2)t^2.

H(2) = 2u -2g = 79.......(1).

H(8) = 8u -32g = 81......(2).

16eq(1) - eq(2) gives 16(2u-2g)-(8u-32g) = 79*16-81 = 1183

24u = 1055

u = 105/24 = 49.2917 m/s

4(1) - (2) gives 4(2u-2g)-(8u-32g) = 79*4-81= 235

24g = 235

g = 235/24 = 9.7917 m/s^2.

Therefore , H(t) = 49.2917t - (1/2)9.7917t^2 is the equation.

From this , when H(t) = 0. Or 49.2917t - (1/2)9.7917 t^2 = 0

t(49.2917 - (1/2)(9.7917)t) = 0

So t = 0,  or

49.2917 - (1/2)9.7917t = 0. or t = 2*49.2917/9.7917 = 10.068 secs approximately.

Hope this helps.

 

 

 

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