Question

A ball is thrown up with the speed of 0.5m/s. (A)how high will it go before it begins to fall?(B) how long will it take to reach the height?

Answer 1

Step-by-step explanation:

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Answer 2

Given:

• Initial velocity, u = 0.5 m/s
• Since the ball reaches up, final velocity there, v = 0
• Acceleration = a = g (because it is under action due to gravitational force of earth).

To find:

• (a)distance covered by the ball before it begins to fall.
• (b) Time taken by the ball to reach the height.

Solution:

• (a) We know all three equations of kinematics.
• To find height, we are using the third equation of kinematics, which is given by, $v^2 = u^2+2as$
• 0 = $(0.5)^2+2*g*h$  = 0.25*-2gs ( "-" sign because the the ball is travelling back in the opposite direction)
• 0.25 = 2gs
• s = 0.25/2g = 0.25/19.6 = 0.012m = 1.2 cm
• (b) To find time we have a formula, t = u/g
• t = 0.5/9.8 = 0.051 s = 52 ms.

(a) s = 1.2 cm

(b) = 0.051 s = 51 ms