Question

A ball is thrown upwards from the surface of the
moon with a velocity of 19.6 m/s. (a) How much
time will it take to attain the maximum height? (b)
How high will it go?
​

Answer 1

Answer:

a. 24.5 second

b. 120.05 metre

Explanation:

We know that gravity of moon is 1/6 of gravity of earth so by using simple kinemetic formulas here the answers

Answer 2

Given :-

Velocity of the ball thrown upwards from the surface of the moon = 19.6 m/s

To Find :-

Time taken to attain the maximum height.

Maximum height reached by the ball.

Analysis :-

Here we are given with the final velocity and the initial velocity of a ball throws upwards from the surface of the  moon.

Firstly, using the first law of motion substitute the values given accordingly and find the time taken.

Then using the second law of motion substitute the values we got and find the distance accordingly.

Solution :-

We know that,

• t = Time
• u = Initial velocity
• a = Acceleration
• v = Final velocity
• s = Displacement

Using the formula,

$\underline{\boxed{\sf First \ equation \ of \ motion=v=u+at}}$

Given that,

Final velocity (v) = 0 m/s

Initial velocity (u) = 19.6 m/s

Substituting their values,

⇒ 0 = 19.6 + (9.8/6) t

⇒ t = (19.6×6) / 9.8

⇒ t = 117.6 / 9.8

⇒ t = 12 sec

Therefore, time taken to attain the maximum height is 12 sec.

Using the formula,

$\underline{\boxed{\sf Second \ equation \ of \ motion=s=ut+\dfrac{1}{2} at^2}}$

Given that,

Initial velocity (u) = 19.6 m/s

Time (t) = 12 sec

Substituting their values,

⇒ s = (19.6 × 12) + (-9.8/6 × 12 × 12) / 2

⇒ s = 235.2 + (-9.8/6 × 144)/2

⇒ s = 235/2 + (-235.2)/2

⇒ s = 235.2 + -117.6

⇒ s = 117.6 m

Therefore, the the ball will go 117.6 m high.