Baseball Star Bryan is standing at the top of the Sears Tower in Chicago and decides to throw a baseball up with a velocity of 5 m/s. The Sears Tower is 442 meters tall and gravity exerts a constant acceleration of -9.8 m/s/s. Ignore ball mass and wind resistance.
a.) Find equations for acceleration, velocity, and position.
b.) At what time does the ball reach its highest point? How high is that point?
c.) When does the ball hit the ground? How fast is the ball traveling when it hits the ground?
a = -9.8
then v = -9.8t + c
when t = 0 , v = 5 m/s
5 = 0 + c, ----> c = 5
v = -9.8t + 5
s = -4.9t^2 + 5t + k
when t = 0, s =442
442 = 0 + 0 + k
s = -4.8t^2 + 5t + 442
b) highest point when v = 0
0 = -9.8t + 5
t = 5/9.8 = appr .51 sec
when t = .51
s = -4.9(.51)^2 + 5(.51) + 442 = 443.28 m
c) when it hits the ground, s = 0
-4.9t^2 + 5t + 442 = 0
use the quadratic formula to solve this equation
One answer will be negative, of course you will have to reject that one.
sub the positive value of t into your velocity equation.
a = -g = -9.8 m/s^2 The statement gives you the answer.
v = integral a dt = -9.8 t + c
at t = 0
v = Vi = 5 m/s
so c = 5
v = -9.8 t + 5
h = integral v dt = 5 t -(9.8)/2 t^2 + cc
when t = 0
h = Hi = 442
so cc = 442
h = 442 + 5 t - 4.9 t^2 (check your physics book :)
at top, v = 0
0 = 5 - 9.8 t
solve for t
0 = 442 + 5 t - 4.9 t^2
solve quadratic for t when h = 0 the ground
then as usual v = 5 - 9.8 t
a) To find the equations for acceleration, velocity, and position, we can use the equations of motion. These equations are derived from calculus and can be used to describe the motion of an object under constant acceleration.
Let's assume that the initial position of the ball is 0 and the direction of motion is upwards. In this case, the acceleration due to gravity is -9.8 m/s² (negative because it is acting in the opposite direction of motion).
The position of the ball can be described using the equation:
s = s₀ + v₀t + (1/2)at²
Here, s represents the position of the ball at time t, s₀ is the initial position, v₀ is the initial velocity, a is the acceleration, and t is the time.
The velocity of the ball at any given time can be calculated using the equation:
v = v₀ + at
And the acceleration is constant at -9.8 m/s².
b) To find the time when the ball reaches its highest point, we know that at its highest point, the velocity will be zero. Using the equation v = v₀ + at, we can set v = 0 to find the time (t) it takes for the velocity to become zero. Substituting the known values, we have:
0 = 5 - 9.8t
Solving for t:
9.8t = 5
t ≈ 0.51 seconds
To find the height of the highest point, we can substitute this value of t into the equation for position:
s = s₀ + v₀t + (1/2)at²
At the highest point, the position will be the height of the Sears Tower, which is 442 meters. Substituting the known values, we have:
442 = 0 + 5(0.51) + (1/2)(-9.8)(0.51)²
Solving for the height:
442 = 2.55 - 0.129
The height at the highest point is approximately 2.42 meters.
c) The time it takes for the ball to hit the ground can be found by setting the position to be equal to the initial height of the Sears Tower (442 m) and solving for t:
442 = 0 + 5t + (1/2)(-9.8)t²
This equation is a quadratic equation, which can be solved to find the time it takes for the ball to hit the ground. Let's solve it using the quadratic formula:
t = (-b ± sqrt(b² - 4ac)) / (2a)
Here, a = (1/2)(-9.8), b = 5, and c = -442. Substituting these values into the quadratic formula:
t = (-5 ± sqrt(5² - 4(1/2)(-9.8)(-442))) / (2(1/2)(-9.8))
Simplifying:
t = (-5 ± sqrt(25 - 4(1/2)(-9.8)(-442))) / (-9.8)
t ≈ (-5 ± sqrt(25 + 2156.8)) / (-9.8)
Now, calculating the square root:
t ≈ (-5 ± sqrt(2181.8)) / (-9.8)
Using a calculator, we find two possible values for t:
t ≈ -21.5 seconds (not relevant in this case)
t ≈ 44.8 seconds
Thus, the time it takes for the ball to hit the ground is approximately 44.8 seconds.
To find the speed of the ball when it hits the ground, we can substitute this time value into the equation for velocity:
v = v₀ + at = 5 - 9.8(44.8)
Calculating the value:
v ≈ -439.04 m/s
The negative sign indicates that the ball is moving downward. Therefore, the ball is traveling downward with a speed of approximately 439.04 m/s when it hits the ground.