If an object is thrown in an upward direction from the top of a building 160 ft high at an initial speed of 30 mi/h, what is its final speed when it hits the ground?
when does it hit?
30mph = 44ft/s
160+44t-16t^2 = 0
find t then plug that into
v(t) = 44-32t
To find the final speed of the object when it hits the ground, we need to take into account the initial speed, the height, and the acceleration due to gravity.
Here's how we can approach this problem step by step:
Step 1: Convert the initial speed from mph to ft/s.
- To convert from mph to ft/s, you can multiply the value by 1.467.
- In this case, the initial speed is 30 mi/h, so the conversion is:
initial_speed_fts = 30 mi/h * 1.467 ft/s = 44.01 ft/s.
Step 2: Use the kinematic equation to find the time it takes for the object to reach the ground.
- We will use the equation h = vi*t + (1/2)*a*t^2, where:
h is the height (160 ft),
vi is the initial velocity (44.01 ft/s),
a is the acceleration due to gravity (-32.17 ft/s^2),
and t is the time we want to find.
- Since the object is thrown upwards, the initial velocity is positive and the acceleration due to gravity is negative (-32.17 ft/s^2).
- Plugging in the given values, we get:
160 ft = 44.01 ft/s * t + (1/2)*(-32.17 ft/s^2)* t^2.
- Rearranging the equation to standard quadratic form (to solve for t), we get:
-16.085 t^2 + 44.01 t - 160 = 0.
- Solving this quadratic equation will give us two values of t. We take the positive value as the object's time of flight since time can't be negative.
Step 3: Calculate the final speed when hitting the ground.
- Once we find the time it takes for the object to reach the ground, we can use the equation vf = vi + a*t to find the final speed (vf).
- The initial velocity (vi) is 44.01 ft/s, the acceleration (a) is -32.17 ft/s^2, and we found the time (t) in the previous step.
- Plugging in the values, we have:
vf = 44.01 ft/s + (-32.17 ft/s^2) * t.
With these steps, we can solve the problem and find the final speed when the object hits the ground.