Question

a ball is thrown vertically upwards and reaches a maximum height of 15m. Calculate the velocity with which the ball was thrown upwards

help​

Answer 1

Answer:

17.15 m/s

Explanation:

${vf}^{2} = {vi}^{2} - 2ad$

When the object reached the Maximum Height, The Final velocity is at zero

$0 = {vi}^{2} - 2ad$

$2ad = {vi}^{2}$

$d = yf - yi$

- The Final position is the Maximum Height

- Let your acceleration be equal to gravity which is 9.8 m/s^2

- The initial position is equal to Zero

$2(9.8)(15 - 0)= {vi}^{2}$

$\sqrt{ {vi}^{2} } = \sqrt{2(9.8)(15)}$

The initial velocity will indicate the velocity in which the ball thrown up

$vi = 17.15 \: \frac{m}{ {s}}$

Answer 2

Answer:

A ball is thrown vertically upwards and reaches a maximum height of 15m.  The velocity with which the ball was thrown upwards is $17.3205\ m/s$

Explanation:

We can solve this problem by using the third law of motion.

• If v is the final velocity of the object, u is the initial velocity of the object, a is the acceleration and s is the distance covered by the object.

• Then, according to the third law of motion,

        $v^{2} - u^{2} =2as$    ...(1)

       

In the given problem,

$s = h_{max}$  

$a =- g$

(since the ball is thrown upward)

where g is the acceleration due to gravity and $h_{max}$ is the maximum height covered by the ball.

$v = 0$

$g = -10\ m/s^2$

$h_{max} =15\ m$

$u = ?$

Substitute these values into equation(1)

$0^{2} - u^{2} =2\times-10\times 15$

$u = \sqrt{300} = 17.3205\ m/s$