Question

a) find the kumet takes bcfore the stone Anather Stone as thrn uertica

Answer 1

Answer:

(a) Given that t = 0 s and initial speed u = 20 m/s

using the kinematics equation v = u + at,

In the upward motion a = -g then v = u - gt, At maximum height the speed of the stone becomes zero, that means final speed is zero i.e., v = 0

then 0 = u - gt or u = gt   or t = u / g and let us assume this as t₁ = u/g

and using the expression s = ut + 1/2gt², for free fall body u = 0 then s = 1/2gt² and using the value t = u/g then

s = 1/2 g(u/g)² = u²/2g = maximum height H,

After reaching maximum height, the stone begins to travel downward like a free fall object. So height h = 1/2 gt₂² or

t₂² = 2h/g = 2 (u²/2g)/g = u²/g², therefore t₂ = u/g

From this we can say that the time of ascent is equal to the time of descent

Therefore, total time t = t₁ + t₂ = u/g + u/g = 2u/g

Then t = 2 (20 m/s) / (9.8 m/s²) = 4.08 s

(b) Maximum height reached by the stone is h_max = u²/2g = (20 m/s)²/ 2 (9.8 m/s²) = 20.40 m.

(c) Two stones thrown vertically upwards with the same speed 20 m/s but with the time difference 2 s

then using the expressions , s₁ = ut - 1/2 gt² and s₂ = u(t-2) - 1/2 g(t-2)²

Equating these two equations then we obtain

ut - 1/2 gt^{2 =} u(t-2) - 1/2 g(t-2)²

2u = 1/2 gt² - 1/2 g(t-2)² = g/2 [ t² - (t-2)² ] = g/2 [ t² - (t² + 4 - 4t) ] = g/2 [4t - 4] = g (2t - 2)

Therefore, 2u = g (2t - 2) or u = g/2 (2t - 2)

Substituting the values then we get,

20 m/s = (9.8 m/s²)/2 (2t - 2)

4.08 s = 2t - 2 or t = 3.04 s at this time the two stones will meet.

using the equation, s₁ = ut - 1/2 gt²

height s₁ = (20 m/s) (3.04 s) - 1/2 (9.8 m/s²) (3.04 s)² = 60.81 m - 45.28 m = 16.53 m at this point the two stones will meet.