Question

A stone is dropped from the top of a tree cover 6m height in last 0.2 s of fall.Find total height of tree?

Answer 1

Answer:

48.05 m

Explanation:

let velocity at that point be v

v = $\sqrt{2gh}$

6 = v(0.2) + 0.5(g)(0.2)(0.2)

6 = 0.2v + 0.2

30 = v + 1

v = 29

29*29 = 2*10*h

h = 42.05

total height = 42.05 + 6 = 48.05 m

Answer 2

The answer is 8.45 meters.

GIVEN

A stone is dropped from the top of a tree cover 6m height in last 0.2 s of fall.

TO FIND

The total height of the tree.

SOLUTION

We can simply solve the above problem as follows;

Let the total height of the tree = H meters

Total time taken by the ball to fall = T

Applying second of motion;

H = uT + 1/2 gT² (u = 0m/s free fall)

H = (1/2)×gT² (Equation 1)

It is given;

In Last 0.2 seconds the ball covers a distance of 6 meters

So,

H-6 = 1/2 × g × (T-0.2)² (Equation 2)

From equation 1 and 2

$\frac{1}{2} gt^{2} - 6 = \frac{1}{2} g(t - 2)^{2}$

= (1/2)× g× (2T-2)× 2 = (1/2)× g× (T-2)²

T = 1.3 seconds

Putting the value of T in (Equation 1)

H = (1/2) × 10 × 1.3²

= 8.45meters

Hence, The answer is 8.45 meters.

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