A boy runs off the edge of a 5 foot high wall @ 2.5 ft/sec. How long will he spend in the air? How far from the base of the wall will he land?
time to fall 5 feet?
hf=ho+vivert*t-16t^2
-5=0+0-16 t^2
solve for time
how far?
distnace=2.5*time in feet.
To find the time the boy spends in the air, we can use the equation of motion:
h = ut + (1/2)gt^2
Where:
- h is the height of the wall (5 ft).
- u is the initial vertical velocity (since the boy runs off the edge, it is 0 ft/sec).
- g is the acceleration due to gravity (-32 ft/sec^2).
- t is the time in seconds.
Substituting the given values, we get:
5 = 0t + (1/2)(-32)t^2
Simplifying, we have:
5 = -16t^2
Dividing both sides by -16, we get:
t^2 = -5/16
Since time cannot be negative, this means that the boy will not spend any time in the air. However, we can still find the horizontal distance the boy will travel before landing.
The horizontal distance covered can be calculated using the formula:
d = vt
Where:
- d is the horizontal distance.
- v is the horizontal velocity (2.5 ft/sec).
- t is the time in seconds.
Plugging in the values, we have:
d = 2.5t
Since we found that the boy spends no time in the air, the value of t is 0. Therefore, the horizontal distance (d) is also 0.
Hence, the boy will spend no time in the air and will land directly at the base of the wall, resulting in no horizontal distance traveled.