Question

a golf ball is released from the rest from top of a very tall building.calculate the position of m of the ball after 2.00 seconds.

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Answer 1

Answer:

Given :

v0 = 0

g = -9.81

t 1,2,3

now we use v = v0 + at for each time

v = -9.81 X (1) =-9.81

v = -9.81 X (2) = -19.6

v = -9.81 X (3) = -29.7

Explanation:

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Answer 2

Answer:

After 2 second ball will be at 19.6 m below the top of the building.

Explanation:

• The ball is falling from a height where it was at rest. So initial velocity of the ball is, u = 0.
• The ball is falling due to gravity and in downward direction. So the acceleration of the ball(a)= acceleration due to gravity, g= 9.8 m/s²
• Time, t = 2 s.
• Suppose in 2s the ball has crossed a distance 'h'.
• From the 2nd equation of motion, we know s = ut + $\frac{1}{2}$at²
• Calculation of height: Substituting all the values in above equation,

                                                h = (0×2) + ($\frac{1}{2}$ × 9.8 × 2²)

                                           ⇒ h = 0 + (9.8 × 2)

                                           ⇒ h = 19.6 m

∴The ball's position after 2 second is: 19.6 m below the top of the building.