Question

find the speed with which a ball can be thrown upwads to reach a height of 10m​

Answer 1

Let the ball is thrown upward with u as initial velocity. As we know that as the ball reaches a maximum height, its final velocity becomes 0.

Sign Convention -

• '+' for upward direction
• '-' for downward direction

Now,

• Initial velocity = u m/s
• Final velocity, v = 0 m/s
• Distance, s = + 10 m
• Acceleration, a = - 9.8 m/s² (as it always apply in downward direction)

Formula to be used = v² - u² = 2as

⇒ (0)² - u² = 2(- 9.8)(10)

⇒ - u² = - 196

⇒ u² = 196

⇒ u = √196

⇒ u = 14

Hence, 14 m/s is the initial velocity.

Answer 2

SOLUTION :-

Taking g = 9.8 m/s²

We need to find the initial velocity u at which a ball should be thrown to reach a height s = 10 m.

At the highest point which is s here , The final velocity v will be 0 m/s.

Using third equation of motion, we have

⇒ 2(-g)s = v² - u²

⇒ 2×-9.8×10 = 0 - u²

⇒ u² = 196

⇒ u = 14

Hence, The ball should be thrown at 14 m/s.

Taking g = 10 m/s²

Now, As we know, Final velocity would be 0 m/s and let the initial velocity be u,

Using third Equation of motion,

⇒ 2(-g)s = v² - u²

⇒ 2×-10×10 = 0 - u²

⇒ u² = 200

⇒ u = 10√2

∴ The ball should be thrown at 10√2 m/s.

Well, There is also a simpler way to solve the given problem. (I would call it the shortcut method)

If air resistance is negligible (Which is indeed negligible here)

The shortcut is:

⇒ Maximum height reached = u² / 2g

Where,

• u = Initial velocity
• g = Acceleration due to gravity

Now, It is given that maximum height reached is 10m.

⇒ h = u² / 2g

⇒ 10 = u² / 2×10

⇒ u² = 200

⇒ u = 10√2 m/s

Now, Taking g = 9.8 m/s²

⇒ h = u² / 2g

⇒ 10 = u² / 2×9.8

⇒ u² = 98×2

⇒ u = 14 m/s

Which is same im both the cases. But since you haven't defined the value of g, So We would have to take both -10m/s² & -9.8m/s² as well.