Question

A ball is thrown vertically upwards and rises to a height of 19.6 m. Calculate
(0) The initial velocity of the ball
(ii) Time taken by the ball to reach the maximum height.
Given g = 9.8 m/s2​

Answer 1

Answer:

(i) Initial velocity is 19.6 m/s.

(ii) Time taken is 2 seconds.

Explanation:

Given,

• Height/Distance, s is 19.6 m.
• Final velocity, v is 0 m/s.
• Acceleration due to gravity, g or a is 9.8 m/s or a = -9.8 m/s [Upward].

(i) We know,

Third equation of motion :

• v² = u² + 2as

[Where, v is final velocity, u is initial velocity, a is acceleration and s is distance/displacement].

Put all values:

$\longrightarrow$ (0)² = (u)² + 2 × (-9.8) × 19.6

$\longrightarrow$ 0 = u² - 19.6 × 19.6

$\longrightarrow$ 0 = u² - (19.6)²

$\longrightarrow$ (19.6)² = u²

• Square root both side

$\longrightarrow$ √(19.6)² = √u²

$\longrightarrow$ u = 19.6

Thus,

Initial velocity of the ball is 19.6 m/s .

(ii) We know,

First equation of motion

• v = u + at

[Where, t is time]

Put all values :

$\longrightarrow$ (0) = 19.6 + (-9.8) × t

$\longrightarrow$ 0 - 19.6 = -9.8 × t

$\longrightarrow$ (-19.6)/(-9.8) = t

$\longrightarrow$ t = 2

Thus,

Time taken by the ball to reach the maximum height is 2 seconds.

Answer 2

Given :-

• Final velocity of ball (v) = 0 m/s
• Distance covered by ball (s) = height = 19.6 m
• Acceleration of ball (a) = g = 9.8 m/s²

To Find :-

• Initial velocity of ball (u)?
• Time taken by ball (t)?

Solution :-

Using third eqⁿ of motion. We know that;

$\hookrightarrow\:{\large{\boxed{\sf{v^2 = u^2 + 2as}}}}$

As ball is thrown upward. So, acceleration will be -9.8 m/s²; (sign is negative because acceleration is opposite to the motion of ball).

Putting all values :-

$\longmapsto\:\tt 0^2 = u^2 + 2(-9.8)(19.6)$

$\longmapsto\:\tt 0 = u^2 + (2\:\times\:-9.8\:\times\:19.6)$

$\longmapsto\:\tt 0 = u^2 + (-384.16)$

$\longmapsto\:\tt 0 = u^2 - 384.16$

$\longmapsto\:\tt 0 + 384.16 = u^2$

$\longmapsto\:\tt u^2 = 384.16$

$\longmapsto\:\tt u = \sqrt{384.16}$

$\longmapsto\:\tt u = \sqrt{19.6\:\times\:19.6}$

$\purple{\longmapsto}\:{\large{\underline{\boxed{\bf{\blue{u} \green{=} \red{19.6}\:\pink{m/s}}}}}}$

Hence, initial velocity of ball is 19.6 m/s.

Using first eqⁿ of motion. We know that;

$\hookrightarrow\:{\large{\boxed{\sf{v = u + at}}}}$

As ball is thrown upward. So, acceleration will be -9.8 m/s²; (sign is negative because acceleration is opposite to the motion of ball).

Putting all values :-

$\longmapsto\:\tt 0 = 19.6 + (-9.8)t$

$\longmapsto\:\tt 0 - 19.6 = -9.8t$

$\longmapsto\:\tt -19.6 = -9.8t$

$\longmapsto\:\tt -9.8t = -19.6$

$\longmapsto\:\tt t = \dfrac{\cancel{-}19.6}{\cancel{-}9.8}$

$\longmapsto\:\tt t = {\cancel{\dfrac{19.6}{9.8}}}$

$\purple{\longmapsto}\:{\large{\underline{\boxed{\bf{\blue{t} \green{=} \red{2}\:\pink{s}}}}}}$

Hence, time taken by ball to reach the maximum height is 2 seconds.

Learn More :-

Three equations of motion,

1. v = u + at
2. s = ut + ½ at²
3. v² = u² + 2as

Where,

• v denotes final velocity
• u denotes initial velocity
• a denotes acceleration
• t denotes time taken
• s denotes distance covered

Some important definitions,

• Acceleration

Acceleration is the process where velocity changes. Since, velocity is the speed and it has some direction. So, change in velocity is considered as acceleration.

• Initial velocity

Initial velocity is the velocity of the object before the effect of acceleration.

• Final velocity

After the effect of acceleration, velocity of the object changes. The new velocity gained by the object is known as final velocity.

• Distance covered

The total path length travelled by an object is known as distance covered.

▬▬▬▬▬▬▬▬▬▬▬▬