Question

A stone is thrown vertically upwards with a speed of 30m/s. How high will it go before it begins to fall. Acceleration is 10m/s square in the downward direction.​

Answer 1

Given :

▪ Initial speed = 30mps

▪ Acc. due to gravity = 10m/s²

To Find :

▪ Maximum height reached by ball.

Concept :

↗ Since, acceleration due to gravity has constant value, we can easily apply equation of kinematics to solve this question.

↗ For a body thrown vertically upward, g is taken negative.

↗ Velocity at highest point = zero

Calculation :

$\dashrightarrow\sf\:v^2-u^2=2(-g)H\\ \\ \dashrightarrow\sf\:(0)^2-(30)^2=-2(10)H\\ \\ \dashrightarrow\sf\:H=\dfrac{900}{20}\\ \\ \dashrightarrow\underline{\boxed{\bf{\red{H=45m}}}}\:\orange{\bigstar}$

Answer 2

$\sf{\underline{Answer\::}}$

The stone will attain a height of 30 m before reaching the point of fall with initial velocity of 30 m/s.

$\sf{\underline{Explanation\::}}$

Given that the stone is thrown vertically upwards, which means against the gravity, therefore value of acceleration due to gravity will be negative.

Also,the intial velocity (u) is 30 m/s.

The value of acceleration due to gravity (g) is 10 m/s².

We are supposed to find the height that the stone will be attaining with a speed of 30 m/s before it reaches the point from where it would start falling down.

Using the third equation of motion,

• v² = u² + 2as

Plug in the values,

=> $\sf{0^2 = 30^2 + 2(-10)s}$

=> $\sf{0 = 900 + (-20)s}$

=> $\sf{0 = 900 - 20s}$

=> $\sf{0 - 900 = - 20s}$

=> $\sf{-900 = - 20s}$

=> $\sf{\dfrac{-900}{-20}=s}$

=> $\sf{\dfrac{90}{2}=s}$

=> $\sf{45 =s}$

•°• The stone will start falling down after reaching a height of 45 m.