You throw a 1kg ball out of a window 12 meters above the ground with horizontal speed. It lands on the ground 5 meters from the building after 1 second. Ignore air resistance, A) How fast did you throw the ball? B) What was the ball's initial velocity? C) How fast was it traveling when it hits the ground?
There is no reason to post the same question more than once.
To solve this problem, we can use the principles of kinematics, which is the study of motion. We'll need to break down the given information and use the appropriate equations to find the answers to each part of the question.
A) How fast did you throw the ball?
To find the vertical velocity when the ball hits the ground, we'll start by using the equation for vertical motion:
y = y0 + (v0y * t) + (0.5 * a * t^2)
In this equation:
- y is the vertical displacement (y = -12m, since the ball is thrown from a height of 12m above the ground).
- y0 is the initial vertical position (y0 = 0, since the ball starts from the ground).
- v0y is the vertical component of the initial velocity (which we need to find).
- t is the time taken for the ball to land (t = 1s).
- a is the acceleration due to gravity (a = -9.8m/s^2, taking downward as the positive direction).
By substituting the known values into the equation, we can solve for v0y:
-12 = 0 + (v0y * 1) + (0.5 * -9.8 * 1^2)
Simplifying this equation gives:
-12 = v0y - 4.9
Rearranging and solving for v0y:
v0y = -12 + 4.9
v0y = -7.1 m/s
Therefore, the vertical component of the initial velocity is -7.1 m/s.
B) What was the ball's initial velocity?
To find the initial velocity of the ball (assuming horizontally thrown), we'll use the equation for horizontal motion:
x = x0 + (v0x * t)
In this equation:
- x is the horizontal displacement (x = 5m, the distance the ball lands from the building).
- x0 is the initial horizontal position (x0 = 0).
- v0x is the horizontal component of the initial velocity (which we need to find).
- t is the time taken for the ball to land (t = 1s).
Since the ball is thrown horizontally, the initial horizontal velocity (v0x) remains constant throughout the motion. Therefore, we can directly determine it from the equation:
5 = 0 + (v0x * 1)
Simplifying, we find:
v0x = 5 m/s
Hence, the ball's initial velocity in the horizontal direction is 5 m/s.
C) How fast was it traveling when it hits the ground?
To find the overall speed of the ball when it hits the ground, we can use the Pythagorean theorem. The initial velocity can be treated as a vector with horizontal and vertical components. The overall speed is equivalent to the magnitude of the initial velocity.
Applying the Pythagorean theorem, we have:
v^2 = v0x^2 + v0y^2
Substituting the values we found earlier, we get:
v^2 = (5^2) + (-7.1^2)
v^2 = 25 + 50.41
v^2 = 75.41
Taking the square root of both sides:
v ≈ 8.68 m/s
Therefore, the ball was traveling at approximately 8.68 m/s when it hit the ground.