An archer shoots an arrow horizontally with an initial velocity of 100 meters per second from a building that is 20 meters high. (air resistance is neglected)
a. Calculate the time when the arrow is in the air.
b. Find the horizontal distance the arrow travels before striking the level ground.
To solve this problem, we can use the equations of motion for projectile motion. We'll assume that the acceleration due to gravity is approximately 9.8 m/s^2, and neglect air resistance.
a. To find the time when the arrow is in the air, we can use the equation for vertical motion:
y = v0y * t + (1/2) * g * t^2
where:
y = vertical displacement (20 meters)
v0y = initial vertical velocity (0 m/s since the arrow is shot horizontally)
g = acceleration due to gravity (-9.8 m/s^2)
t = time
Putting the values into the equation, we get:
20 = 0 * t + (1/2) * (-9.8) * t^2
20 = -4.9t^2
Rearranging the equation, we have:
4.9t^2 = -20
t^2 = -20 / 4.9
t^2 = 4.08
t ≈ √4.08
t ≈ 2.02 seconds
Therefore, the arrow is in the air for approximately 2.02 seconds.
b. To find the horizontal distance the arrow travels before striking the ground, we can use the equation for horizontal motion:
x = v0x * t
where:
x = horizontal distance
v0x = initial horizontal velocity (100 m/s)
t = time (2.02 seconds)
Plugging in the values, we get:
x = 100 * 2.02
x ≈ 202 meters
Therefore, the arrow travels approximately 202 meters horizontally before striking the level ground.
a. To calculate the time when the arrow is in the air, we can use the formula for vertical motion:
h = ut + (1/2)gt^2
where:
h = height (20 meters in this case)
u = initial vertical velocity (0 meters per second since shot horizontally)
g = acceleration due to gravity (approximately 9.8 meters per second squared)
t = time
Since the initial vertical velocity is 0 (as the arrow is shot horizontally), the equation simplifies to:
h = (1/2)gt^2
Plugging in the values, we have:
20 = (1/2)(9.8)t^2
Simplifying further:
20 = 4.9t^2
Dividing both sides by 4.9:
4.08 = t^2
Taking the square root of both sides:
t ≈ 2.02 seconds
Therefore, the time when the arrow is in the air is approximately 2.02 seconds.
b. To find the horizontal distance the arrow travels before striking the level ground, we can use the formula for horizontal motion:
d = ut
where:
d = horizontal distance
u = initial horizontal velocity (100 meters per second in this case)
t = time
Plugging in the values, we have:
d = 100 × 2.02
d ≈ 202 meters
Therefore, the horizontal distance the arrow travels before striking the level ground is approximately 202 meters.
a. d = Vo*t + 0.5g*t^2 = 20 m.
0 + 4.9t^2 = 20,
t^2 = 4.08,
t = Tf = 2.02 s. = Fall time = Time in
flight.
b. Dx=Xo * Tf=100 m/s * 2.o2 s.=202 m.