two stones are thrown vertically up at the same time. One reaches a height 35 m more than the other and touches the ground 2 s later. What are the initial velocities of the stones?
Answers
Answer:
30 & 40 m/s
Explanation:
Let say first stone Velocity = V₁
and it takes 2T₁ time to reach back to ground ( T₁ to go up & T₁ to come back)
height H₁
a = g (- during upward)
Velocity at top = 0
0 = V₁ - gT₁
=> T₁ = V₁/g Eq 1
0² - V₁² = 2(-g)H₁
=> V₁² = 2(g)H₁ Eq 2
Second stone Velocity = V₂
and it takes 2T₁ + 2 time to reach back to ground ( T₁ + 1 to go up & T₁ + 1 to come back)
height H₁ + 35
T₁ + 1 = V₂/g
=> T₁ = V₂/g - 1 Eq3
=> V₂² = 2(g)(H₁ + 35)
=> 2(g)(H₁ = V₂² - 70g Eq 4
Equating eq 1 & Eq 3
V₁/g = V₂/g - 1
=> V₁ = V₂ - g
Equating eq 2 & 4
V₁² = V₂² - 70g
putting V₁ = V₂ - g
=> (V₂ - g)² = V₂² - 70g
=> V₂² + g² - 2V₂g = V₂² - 70g
=> g² - 2V₂g = - 70g
=> g - 2V₂ = -70
=> V₂ = (70 + g)/2
putting g= 10
=> V₂ = 40 m/s
V₁ = V₂ - g = 40 - 10 = 30 m/s
Initial Velocity of stones are 30 & 40 m/s