First, you will investigate purely vertical motion. The kinematics equation for vertical motion (ignoring air resistance) is given by
y(t)=y0+v0t−(1/2)gt2,
where y0=0 is the initial position (which is 1.2 m above the ground due to the wheels of the cannon), v0 is the initial speed, and g is the acceleration due to gravity.
Shoot the baseball straight upward (at an angle of 90∘) with an initial speed of 20 m/s.
How long does it take for the baseball to hit the ground?
Express your answer with the appropriate units.
g = 9.81 m/s^2 so g/2 = 4.9
0 = 1.2 + 20 t - 4.9 t^2
or
4.9 t^2 - 20 t - 1.2 = 0
solve quadratic for t (use positive answer choice)
To find the time it takes for the baseball to hit the ground, we can use the kinematic equation for vertical motion:
y(t) = y0 + v0t - (1/2)gt^2
The initial position, y0, is 1.2 m above the ground due to the wheels of the cannon. Since the baseball is shot straight upward, the initial velocity, v0, is 20 m/s. The acceleration due to gravity, g, is approximately 9.8 m/s^2.
Setting y(t) to 0 (since the baseball hits the ground), we can solve for t:
0 = 1.2 + (20)t - (1/2)(9.8)t^2
Rearranging the equation and setting it equal to zero:
(1/2)(9.8)t^2 - (20)t - 1.2 = 0
Now we can solve this quadratic equation to find the values of t.
Using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
Plugging in the values of a, b, and c:
a = (1/2)(9.8) = 4.9
b = -20
c = -1.2
t = (-(-20) ± √((-20)^2 - 4(4.9)(-1.2))) / (2(4.9))
Simplifying:
t = (20 ± √(400 + 23.52)) / 9.8
t = (20 ± √423.52) / 9.8
Since we are looking for the time it takes for the baseball to hit the ground, we only consider the positive value of t.
t = (20 + √423.52) / 9.8
Now we can use a calculator to find the exact value of t.
To determine the time it takes for the baseball to hit the ground, we need to consider the vertical motion of the baseball. In this case, the baseball is shot straight upward (at an angle of 90 degrees) with an initial speed of 20 m/s.
The kinematics equation for vertical motion is given by:
y(t) = y0 + v0t - (1/2)gt^2
In this equation:
- y(t) represents the vertical displacement at time t
- y0 is the initial vertical position, which is 1.2 m above the ground
- v0 is the initial vertical velocity, which is 20 m/s
- g is the acceleration due to gravity, which is approximately 9.8 m/s^2
Since the baseball is shot straight upward, its initial vertical velocity is positive (upward). Therefore, the equation becomes:
y(t) = 1.2 + 20t - (1/2)(9.8)t^2
Now, we need to find the time it takes for the baseball to hit the ground, which means finding the time when y(t) = 0 (since the ground is at a height of 0).
Setting y(t) = 0, the equation becomes:
0 = 1.2 + 20t - (1/2)(9.8)t^2
To solve for t, we rearrange the equation and set it equal to zero (quadratic equation form):
(1/2)(9.8)t^2 - 20t - 1.2 = 0
Solving this quadratic equation will give us the values of t. We can use the quadratic formula:
t = (-b ± √(b^2 - 4ac))/(2a)
In this case, a = (1/2)(9.8) = 4.9, b = -20, and c = -1.2
Plugging these values into the quadratic formula, we can compute the two values of t. One of the solutions will be negative, so we can discard it (since we're looking for the time it takes for the ball to hit the ground).
Once we solve for t, we'll have the time it takes for the baseball to hit the ground, expressed in the appropriate units (seconds).
Time = -0.059s and 4.14s
we have to take positive number because time never be negative TIME = 4.14s ( coorect answer in my assignment )