An archer shoots an arrow horizontally at a target 14 m away. The arrow is aimed directly at the center of the target, but it hits 52 cm lower.
1-What was the initial speed of the arrow?
** is this the formula for finding t :
y=h-1/2(gt^2)= 00.52-1/2(9.8t^2)
t=0.3258 s
initial speed= 14/0.3258=43 m/s
is my calculation correct?
I didn't use calculator, but is done correctly.
Your calculation is correct. To find the initial speed of the arrow, you can use the horizontal distance and the time it takes for the arrow to hit the target. From your calculations, you determined that the time it takes the arrow to hit the target is approximately 0.3258 s.
Using the formula y = h - (1/2)gt^2, where y is the vertical displacement, h is the initial height, g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time, you found that the vertical displacement is 0.52 m.
To find the initial speed of the arrow, you can use the equation s = vt, where s is the horizontal distance, v is the initial speed, and t is the time. The horizontal distance is given as 14 m, and the time is approximately 0.3258 s.
Therefore, the initial speed of the arrow is calculated as follows:
v = s/t
v = 14 m / 0.3258 s
v ≈ 43 m/s
So, the initial speed of the arrow is approximately 43 m/s.
Your calculation is almost correct, but there seems to be a mistake in the formula you used to solve for time (t). The correct formula to use in this case is the equation for vertical motion:
y = h + v₀t - (1/2)gt²
Where:
y = vertical displacement (height difference at impact)
h = initial height (the arrow hits 52 cm lower than the center of the target, so h = -0.52 m)
v₀ = initial vertical velocity (we assume it is 0 because the arrow is shot horizontally)
t = time of flight
g = acceleration due to gravity (approximately 9.8 m/s²)
So let's correct the calculation:
y = h + v₀t - (1/2)gt²
-0.52 = 0 + (0)t - (1/2)(9.8)(t²)
Simplifying the equation:
-0.52 = -4.9t²
Now, solve for t:
t² = 0.52 / 4.9
t² ≈ 0.1061
t ≈ √0.1061
t ≈ 0.326 s
Now, to find the initial speed, divide the horizontal distance (14 m) by the time (0.326 s):
Initial speed = 14 / 0.326
Initial speed ≈ 42.9 m/s
So the initial speed of the arrow is approximately 42.9 m/s.