An archer shoots at an angle of 11 degrees from the horizontal at an initial speed of 31.8m/s. Ignoring air resistance, determine
a) the time it will take for the arrow to strike the ground
b) the range of the arrow
vertical problem first
Vi = 31.8 sin 11
v = Vi - 9.81 t
at top v = 0
t at top = (31.8 sin 11) / 9.81
t in air = 2 * t at top = 2 (31.8 sin 11) / 9.81 because parabola symmetric )
then horizontal problem now we know how long in air
U = horizontal speed = 31.8 cos 11
There is NO horizontal force
SO there is no change in horizontal speed (until it hits ground)
so
we know t in air and U
distance = U t
sorry i still dont understand how to calculate it ?
t in air = 2 * t at top = 2 (31.8 sin 11) / 9.81
and
distance = 31.8 cos 11 * t in air
thanks very much
You are welcome.
To determine the time it will take for the arrow to strike the ground and the range of the arrow, we can use the equations of motion.
a) First, we need to find the vertical component of the initial velocity. This can be calculated using the equation:
Vy = V * sin(theta)
where Vy is the vertical component of velocity, V is the initial speed, and theta is the launch angle.
Vy = 31.8 m/s * sin(11 degrees)
= 31.8 m/s * 0.191
Vy = 6.08 m/s (approximately)
Now, we can find the time it will take for the arrow to strike the ground using the equation:
t = (2 * Vy) / g
where t is the time of flight and g is the acceleration due to gravity (approximately 9.8 m/s²).
t = (2 * 6.08 m/s) / 9.8 m/s²
= 1.24 s (approximately)
Therefore, the time it will take for the arrow to strike the ground is approximately 1.24 seconds.
b) To find the range of the arrow, we need to find the horizontal component of the initial velocity. This can be calculated using the equation:
Vx = V * cos(theta)
where Vx is the horizontal component of velocity, V is the initial speed, and theta is the launch angle.
Vx = 31.8 m/s * cos(11 degrees)
= 31.8 m/s * 0.981
Vx = 31.20 m/s (approximately)
Now, we can use the time of flight (t) and the horizontal component of velocity (Vx) to find the range (R) using the equation:
R = Vx * t
R = 31.20 m/s * 1.24 s
= 38.68 m (approximately)
Therefore, the range of the arrow is approximately 38.68 meters.