Question

A ball is kicked horizontally with a speed of 5.0 ms-1 from the roof of a house 3 m high. When will the ball hit the ground?

Answer 1

The ball will hit the ground approximately 0.783 seconds after being kicked off the roof. This time is calculated solely based on the vertical descent due to gravity, independently of the ball’s horizontal velocity.

Step-by-step explanation:

The question you've asked is related to the concepts of projectile motion in Physics. Specifically, to determine when the ball will hit the ground after being kicked horizontally from the roof of a house, we need to consider the vertical motion of the ball because the horizontal component does not affect the time taken to hit the ground. We will use the following kinematic equation for vertical motion, which does not depend on the horizontal speed:

h = ½gt²

where:

• h is the height above the ground (3 m in this case)
• g is the acceleration due to gravity (approximately 9.8 m/s²)
• t is the time in seconds.

Rearranging the equation to solve for t, we get:

t = √(2h/g)

Plugging in the values:

t = √(2*3 m/9.8 m/s²) = √(6/9.8) = √(0.6122) ≈ 0.783 s

So, the ball will hit the ground approximately 0.783 seconds after being kicked.

Answer 2

the time taken for the ball to hit the ground is 0.424 s

Step-by-step explanation:

Given;

velocity of the ball, u = 5 m/s

height of the house which the ball was kicked, h = 3m

Apply kinematic equation;

h = ut + ¹/₂gt²

where;

h is height above ground

u is velocity

g is acceleration due to gravity

t is the time taken for the ball to hit the ground

Substitute the given values and solve for t

3 = 5t + ¹/₂(9.8)t²

3 = 5t + 4.9t²

4.9t² + 5t -3 = 0

a = 4.9, b = 5, c = -3

Solve for t using formula method

`t = (-5 +/-√(5^2-4(4.9*-3)))/(2(4.9)) = (-5+/-(9.154))/(9.8) \\\\t = (-5+9.154)/(9.8) \ or \ (-5-9.154)/(9.8) \\\\t = (4.154)/(9.8) \ or \ (-14.154)/(9.8) \\\\t = 0.424 \ sec \ or -1.444 \ sec\\\\Thus, t = 0.424 \ sec`