Question

Queuetion:-

A ball is thrown vertically upwards with a velocity of 49 m/s. Calculate

(i) the maximum height to which it rises,

(ii) the total time it takes to return to the surface of the​

Answer 1

Answer:

Given Initial velocity of ball, u=49 m/s

Let the maximum height reached and time taken to reach that height be H and t respectively.

Assumption: g=9.8 m/s

Velocity of the ball at maximum height is zero, v=0

v²−u²=2aH

0−(49) ²=2×(−9.8)×H

⟹H=122.5 m

v=u+at

0=49−9.8t

⟹t=5 s

∴ Total time taken by ball to return to the surface, T=2t=10 s

Answer 2

Answer:

Given :-

• A ball is thrown vertically upwards with a velocity of 49 m/s.

To Find :-

• What is the maximum height to which it rises.
• What is the total time taken to return to the surface of the Earth.

Formula Used :-

$\clubsuit$ Third Equation Of Motion Formula :

$\bigstar \: \: \sf\boxed{\bold{\pink{v^2 =\: u^2 + 2gh}}}\: \: \: \bigstar$

where,

• v = Final Velocity
• u = Initial Velocity
• g = Acceleration due to gravity
• h = Height

$\clubsuit$ First Equation Of Motion Formula :

$\bigstar \: \: \sf\boxed{\bold{\pink{v =\: u + gt}}}\: \: \: \bigstar$

where,

• v = Final Velocity
• u = Initial Velocity
• g = Acceleration due to gravity
• t = Time Taken

Solution :-

❒ The maximum height to which it rises :

Given :

• Initial Velocity (u) = 49 m/s
• Final Velocity (v) = 0 m/s
• Acceleration due to gravity (g) = 10 m/s²

According to the question by using the formula we get,

$\implies \bf v^2 =\: u^2 + 2gh$

$\implies \sf (0)^2 =\: (49)^2 + 2(- 10)h$

$\implies \sf (0 \times 0) =\: (49 \times 49) + (- 20)h$

$\implies \sf 0 =\: 2401 - 20h$

$\implies \sf 0 - 2401 =\: -20h$

$\implies \sf {\cancel{-}} 2401 =\: {\cancel{-}} 20h$

$\implies \sf 2401 =\: 20h$

$\implies \sf \dfrac{2401}{20} =\: h$

$\implies \sf 120.05 =\: h$

$\implies \sf\bold{\red{h =\: 120.05\: m}}$

$\therefore$ The maximum height to which it rises is 120.05 m .

❒ The total time taken to return to the surface of the Earth :

Given :

• Final Velocity (v) = 0 m/s
• Initial Velocity (u) = 49 m/s
• Acceleration due to gravity (g) = 10 m/s²

According to the question by using the formula we get,

$\implies \bf v =\: u + gt$

$\implies \sf 0 =\: 49 + (- 10)t$

$\implies \sf 0 =\: 49 - 10t$

$\implies \sf 0 - 49 =\: - 10t$

$\implies \sf {\cancel{-}} 49 =\: {\cancel{-}} 10t$

$\implies \sf 49 =\: 10t$

$\implies \sf \dfrac{49}{10} =\: t$

$\implies \sf 4.9 =\: t$

$\implies \sf\bold{\purple{t =\: 4.9\: seconds}}$

Now, we have to find the total time taken :

So,

$\footnotesize \dashrightarrow \bf Total\: Time\: Taken =\: Time\: of\: ascend + Time\: of\: descend$

$\dashrightarrow \sf Total\: Time\: Taken =\: (4.9 + 4.9)\: seconds$

$\dashrightarrow \sf\bold{\red{Total\: Time\: Taken =\: 9.8\: seconds}}$

$\therefore$ The total time taken to return to the surface of the Earth is 9.8 seconds .