how to solve this problem ?
Attachments:
DhruvkundraiMessi:
please kindly type the question for me pleaseee
a ball is thrown vertically up with velocity u. find its velocity at the moment it reaches half the maximum height. what time it will take to reach that height?
Answers
Answered by
1
Hello Friend,
Here, we will break the problem into two parts: In first part, we will consider the time when the rocket is accelerating. In the second part, we will see what happens when the engine is shut off.
♦ PART 1:
From beginning till the time engines are shut off.
→ Initially, rocket is at rest in ground.
So, Initial velocity = u = 0 m/s
Acceleration a = 25 m/s²
Time t = 5 s
Let final velocity = v
Let height achieved (distance travelled vertically) = h1
→ As motion is uniformly accelerated,
v = u + at
So, v = 0 + 25(5)
So, v = 125 m/s² upwards ---(1)
→ Also,
s = ut + ½ at²
So, here, h1 = 0 + ½ (25)(5)²
So, h1 = 625/2 m
So, h1 = 312.5 metres
♦ PART 2:
From the time engines are shut off till the time the rocket reaches maximum height.
→ Here, Initial Velocity = u = 125 m/s upwards (From (1))
Acceleration = g = 10 m/s² downwards
Let distance travelled vertically = h2
As here the speed will be decreasing, when the rocket reaches maximum height, velocity will be zero.
So, final velocity = v = 0
Here also, motion is uniformly accelerated. But here, g is in downward direction, while velocity is upwards. So, we need to take g as negative in calculations.
Here,
v² = u² + 2as
So, 0² = 125² + 2(-g)(h2)
So, 0 = 125² -(2)(10)(h2)
So, 20 h2 = 125²
So, h2 = (125×125)/20 m
So, h2 = (25×125)/4 m
So, h2 = 3125/4 m
So, h2 = 781.25 metres
♦ Final Answer:
In part 1, the rocket reached a height h1 starting from ground. In part 2, the rocket reached h2 height.
So, total maximum height (H) achieved is the sum of h1 and h2.
So, H = h1 + h2
So, H = 312.5 + 781.25 m
So, H = 1093.75 m
So, H ≈ 1100 metres
→ Thus, the rocket reaches a maximum height of approximately 1100 metres.
Hope it helps.
Purva
@Purvaparmar1405
Brainly.in
Here, we will break the problem into two parts: In first part, we will consider the time when the rocket is accelerating. In the second part, we will see what happens when the engine is shut off.
♦ PART 1:
From beginning till the time engines are shut off.
→ Initially, rocket is at rest in ground.
So, Initial velocity = u = 0 m/s
Acceleration a = 25 m/s²
Time t = 5 s
Let final velocity = v
Let height achieved (distance travelled vertically) = h1
→ As motion is uniformly accelerated,
v = u + at
So, v = 0 + 25(5)
So, v = 125 m/s² upwards ---(1)
→ Also,
s = ut + ½ at²
So, here, h1 = 0 + ½ (25)(5)²
So, h1 = 625/2 m
So, h1 = 312.5 metres
♦ PART 2:
From the time engines are shut off till the time the rocket reaches maximum height.
→ Here, Initial Velocity = u = 125 m/s upwards (From (1))
Acceleration = g = 10 m/s² downwards
Let distance travelled vertically = h2
As here the speed will be decreasing, when the rocket reaches maximum height, velocity will be zero.
So, final velocity = v = 0
Here also, motion is uniformly accelerated. But here, g is in downward direction, while velocity is upwards. So, we need to take g as negative in calculations.
Here,
v² = u² + 2as
So, 0² = 125² + 2(-g)(h2)
So, 0 = 125² -(2)(10)(h2)
So, 20 h2 = 125²
So, h2 = (125×125)/20 m
So, h2 = (25×125)/4 m
So, h2 = 3125/4 m
So, h2 = 781.25 metres
♦ Final Answer:
In part 1, the rocket reached a height h1 starting from ground. In part 2, the rocket reached h2 height.
So, total maximum height (H) achieved is the sum of h1 and h2.
So, H = h1 + h2
So, H = 312.5 + 781.25 m
So, H = 1093.75 m
So, H ≈ 1100 metres
→ Thus, the rocket reaches a maximum height of approximately 1100 metres.
Hope it helps.
Purva
@Purvaparmar1405
Brainly.in
Similar questions