an archer defending a castle is on an 15.5 m high wall. he shoots an arrow straight down at 22.8 m/s
I know this is an easier question but it won’t take the right answer can someone help????
Of course, I can help you with that!
To find the answer, we need to calculate the time it takes for the arrow to reach the ground. We can use the equation of motion, specifically the one that relates the vertical displacement, initial velocity, time, and acceleration.
The equation we'll use is:
h = ut + (1/2)gt^2
Where:
h = vertical displacement (15.5 m, since the archer is shooting straight down from a 15.5 m high wall)
u = initial velocity (22.8 m/s, which is the speed of the arrow)
t = time (unknown, what we're trying to find)
g = acceleration due to gravity (-9.8 m/s^2, which is negative since it's acting in the opposite direction to the arrow's motion)
Plugging in the values we have:
15.5 = (22.8)t + (1/2)(-9.8)t^2
This equation is a quadratic equation since it has a t^2 term. We can rearrange it to solve for t:
(1/2)(-9.8)t^2 + (22.8)t - 15.5 = 0
Now we can solve this quadratic equation either by factoring, using the quadratic formula, or other applicable methods to find the two possible values for t. Since time cannot be negative in this context, we will only consider the positive value as the time it takes for the arrow to reach the ground.
Once we have the time value, we can substitute it back into the equation h = ut + (1/2)gt^2 to find the vertical displacement.
using the "free fall" equation ... h = -1/2 g t^2 + vo t + ho
0 = -1/2 * 9.81 * t^2 - 15.5 t + 22.8
use the quadratic formula to find t ... time for arrow to reach the ground
... 1.09 s
sometimes answers aren't taken due to sig fig problems