A ball is thrown horizontally from the roof of a building 55 m tall and lands 50 m from the base. What was the ball's initial speed?
how long does it take to hit the ground?
h=1/2 g t^2
55=4.9 t^2 solve for time t.
horizontal distance= speed*time
solve for speed.
To find the initial speed of the ball, we can use the fact that the horizontal velocity remains constant throughout the motion.
Let's break down the given information:
- The ball is thrown horizontally, which means there is no vertical initial velocity.
- The height of the building is 55 m.
- The horizontal distance traveled by the ball is 50 m.
Since there is no initial vertical velocity, the only force acting on the ball is the force due to gravity, which causes it to fall vertically downward. The time it takes for the ball to fall from the top of the building to the ground can be found using the equation:
h = (1/2) * g * t^2
Where:
h is the height of the building (55 m)
g is the acceleration due to gravity (approximated as 9.8 m/s^2)
t is the time taken
By substituting the given values into the equation, we can find the time it takes for the ball to fall from the roof of the building:
55 = (1/2) * 9.8 * t^2
Now, let's solve for t:
55 * 2 = 9.8 * t^2
110 = 9.8 * t^2
t^2 = 110 / 9.8
t^2 ≈ 11.22
Taking the square root of both sides:
t ≈ √11.22
t ≈ 3.35 seconds
Now that we know the time it takes for the ball to fall, we can calculate its initial horizontal velocity using the equation:
v = d / t
Where:
v is the initial horizontal velocity
d is the horizontal distance traveled (50 m)
t is the time taken (3.35 seconds)
Substituting the values:
v = 50 / 3.35
v ≈ 14.93 m/s
Therefore, the ball's initial speed was approximately 14.93 m/s.