Question

A ball is launched at an angle of 30° above the horizontal with an initial velocity of 8 m/s. At what time is its vertical velocity 0ms?

Answer 1

ANSWER:

Given:

• Launching Angle $\theta$ = 30°
• Initial Velocity(u) = 8m/s

To Find:

• Time at which vertical velocity is 0

Solution:

We know that, the vertical velocity at Maximum Height is 0.

So, we need to find the time taken to reach the Maximum Height.

We know that,

$:\hookrightarrow\text{Time taken to reach Max Height = Time of ascent}$

And,

$\sf:\hookrightarrow\sf\text{Time of ascent} =\dfrac{u\sin\theta}{g}$

So,

$\sf:\implies\text{Time taken to reach Max Height} =\dfrac{u\sin\theta}{g}$

Here,

⇒ u = 8m/s

⇒ $\theta$ = 30°

⇒ g = 10m/s^2

Hence,

$\sf:\implies\text{Time taken to reach Max Height} =\dfrac{(8)\sin(30^{\circ})}{(9.8)}$

As, sin30° = 1/2,

$\sf:\implies\text{Time taken to reach Max Height} =\dfrac{8\!\!\!/^{\:4}\times\frac{1}{2\!\!\!/}}{10}$

So,

$:\implies\text{Time taken to reach Max Height} =\dfrac{4}{10}$

So,

$\bf{:\implies\text{\bf{Time taken to reach Max Height}} =0.4 sec}$

Therefore, the time at which the vertical velocity of the ball is 0 is 0.4seconds.

Formulae Used:

• Time taken to reach Max Height = Time of ascent
• Time of ascent = (usin$\theta$)/g