A ball is projected horizontally from the top of a building 19.6m high.if the line joining the point of projection to the point where it hits the ground makes an angle of 45° to the horizontal what is the initial velocity of the ball
To find the initial velocity of the ball, we can use the following formula:
v₀ = d / t
where:
v₀ is the initial velocity of the ball,
d is the horizontal distance traveled by the ball, and
t is the time taken for the ball to hit the ground.
First, let's calculate the time taken for the ball to hit the ground. We can use the vertical motion equation:
h = (1/2) * g * t²
where:
h is the height of the building (19.6m),
g is the acceleration due to gravity (approximately 9.8 m/s²), and
t is the time taken for the ball to hit the ground.
Rearranging the equation and substituting known values:
t = √(2h / g)
t = √(2 * 19.6 / 9.8)
t = √4
t = 2 seconds
Now, let's calculate the horizontal distance traveled by the ball. We can use the equation:
d = v₀ * t
Rearranging the equation:
v₀ = d / t
v₀ = d / 2
From the given information, we know that the angle between the horizontal line and the line joining the point of projection to the point where it hits the ground is 45°. The horizontal distance traveled by the ball can be calculated using trigonometry:
sin(45°) = d / 19.6
Simplifying the equation:
d = sin(45°) * 19.6
d = 0.707 * 19.6
d = 13.892 meters (approximately)
Now, we can substitute the calculated d value into the initial velocity equation:
v₀ = 13.892 / 2
v₀ = 6.946 m/s (approximately)
Therefore, the initial velocity of the ball is approximately 6.946 m/s.
You want the ball to land 19.6m from the wall.
So, how long does it take to fall 19.6m?
That would be 2 seconds.
So, what horizontal speed does it take to travel 19.6m in 2s?