Question

A ball is thrown vertically upwards. It returns 6 s later. Calculate the greatest height reached by the ball. Take g= 10 m/s^2

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Answer 1

Answer:

Explanation:

Solution:

• Time will be 3 secs as body will attain its maximum height at this time and then will start falling downward.
• Now, Time, t = 3 secs.
• Final velocity, v= 0 g = 10m/s².
• Initial velocity: v= u+gt 0= u+ (-10)x 3 u= 30 m/s .
• Now, height=? v²-u²= 2gh 0²-30²= 2x(-10)xh

=> Height= 45m.............

Answer 2

Solution:

From the question,

Total time taken is six seconds,i.e.,to go upwards and come downwards.

Time taken in one interval is 3 seconds.

Time taken,t:3seconds

g:10m/s^2

To find:

Height

Now,

Initial velocity of the ball is zero. As,it changed its state of motion from rest to motion.

Using the formula,

$h = ut + \frac{1}{2} gt {}^{2}$

Substituting the values,

$h = (0)(3) + \frac{1}{2} \times 10 \times 3 \times 3 \\ \\ \implies \: h = 5 \times 3 \times 3 \\ \\ \implies \: \boxed{h = 45m}$

The maximum distance ball can reach is 45m.