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V = 40 kmh⁻¹ and T = 9 min.
using s = ut + 1/2at² -h = 19.6 x 6+1/2 x (-9.8) x 62 h = 58.8
For the ball chapped from the top x = 4.9t² ...(i) For the ball thrown upwards 100 – x = 25t – 4.9t² ...(ii) From eq. (i) & (ii), t =4s; x = 78.4 m
For the ball chapped from the top x = 4.9t² …(i) For the ball thrown upwards 100 – x = 25t – 4.9t² …(ii) From eq. (i) & (ii), t =4s; x = 78.4 m
position, x = 25 m velocity = dx/dt = 8t-15, t = 0, v = 0 – 15 = – 15 m/s acceleration, a = dv/dt = 8 ms⁻²
Vₐᵥg = S₁ + S₂/ t₁ + t ₂ = S+S/ S(1/V₁ + 1/V₂) = 2V₁V₂/ V₁ V₂ or 40= 2x30xv₂/V₁+V₂ ⇒V₂ = 60kmh⁻¹
. Two town A and B are connected by a regular bus service with a bus leaving in either direction every T min. A man cycling with a speed of 20 kmh⁻¹ in the direction A to B notices that a bus goes past him every 18 min in the direction of his motion, and every 6 min in the opposite direction. What is the period T of the bus service and with what speed do the busses ply of the road ?
V = 40 kmh⁻¹ and T = 9 min.
V = 40 kmh⁻¹ and T = 9 min.
See lessA ball thrown vertically upwards with a speed of 19.6 ms⁻¹ from the top of a tower returns to the earth in 6s. Find the height of the tower. (g = 9.8 m/ s²)
using s = ut + 1/2at² -h = 19.6 x 6+1/2 x (-9.8) x 62 h = 58.8
using s = ut + 1/2at²
See less-h = 19.6 x 6+1/2 x (-9.8) x 62
h = 58.8
From the top of a tower 100 m in height a ball is dropped and at the same time another ball is projected vertically upwards from the ground with a velocity of 25 m/s. Find when and where the two balls will meet ? (g = 9.8 m/s)
For the ball chapped from the top x = 4.9t² ...(i) For the ball thrown upwards 100 – x = 25t – 4.9t² ...(ii) From eq. (i) & (ii), t =4s; x = 78.4 m
For the ball chapped from the top
See lessx = 4.9t² …(i)
For the ball thrown upwards
100 – x = 25t – 4.9t² …(ii)
From eq. (i) & (ii),
t =4s; x = 78.4 m
The displacement x of a particle varies with time as x = 4t² – 15t + 25. Find the position, velocity and acceleration of the particle at t = 0.
position, x = 25 m velocity = dx/dt = 8t-15, t = 0, v = 0 – 15 = – 15 m/s acceleration, a = dv/dt = 8 ms⁻²
position, x = 25 m
See lessvelocity = dx/dt = 8t-15,
t = 0, v = 0 – 15 = – 15 m/s
acceleration, a = dv/dt = 8 ms⁻²
On a 60 km straight road, a bus travels the first 30 km with a uniform speed of 30 kmh⁻¹. How fast must the bus travel the next 30 km so as to have average speed of 40 kmh⁻¹ for the entire trip ?
Vₐᵥg = S₁ + S₂/ t₁ + t ₂ = S+S/ S(1/V₁ + 1/V₂) = 2V₁V₂/ V₁ V₂ or 40= 2x30xv₂/V₁+V₂ ⇒V₂ = 60kmh⁻¹
Vₐᵥg = S₁ + S₂/ t₁ + t ₂ = S+S/ S(1/V₁ + 1/V₂) = 2V₁V₂/ V₁ V₂
See lessor 40= 2x30xv₂/V₁+V₂ ⇒V₂ = 60kmh⁻¹