An arrow is fired horizontally with a speed of 30ms-1 from a height of 2m above the ground level. Model the arrow as a particle which does not experience air resistance.
a)show that the time taken for the arrow
to hit the ground is 0.639 seconds correct to 3.s.f
b)find the horizontal distance travelled by the arrow
c)describe what happens during the flight of the arrow to the magnitude of:
i)the horizontal component of the velocity
ii)the vertical component of the velocity
please help im very confused give a very clear helpful explanation pplease so i can understand it
a. h = 0.5g*t^2.
h = 2 m, g = 9.8 m/s^2, t = ?.
b. d = Xo*t.
Xo = 30 m/s., t = Value from part a, d = ?.
c.
i. The horizontal component of velocity is initially 30 m/s and remains constant.
ii. The vertical component is initially 0 and increases as the arrow moves toward ground:
Y = Yo + g*t = 0 + 9.8*0.639 = 6.26 m/s. = Max vertical component reached.
a) To find the time taken for the arrow to hit the ground, we need to analyze the vertical motion of the arrow. Since the arrow is fired horizontally, it will not have any initial vertical velocity.
Using the equation of motion:
y = ut + (1/2)gt^2
where:
y = vertical displacement (in this case, the initial height of 2m)
u = initial vertical velocity (which is 0 m/s as it starts horizontally)
g = acceleration due to gravity (-9.8 m/s^2)
t = time taken (unknown)
Substituting the given values:
2 = 0 × t + (1/2) × (-9.8) × t^2
2 = (-4.9) t^2
Dividing both sides by (-4.9):
0.408163 = t^2
Taking the square root:
t ≈ 0.639 seconds (correct to 3 significant figures)
Therefore, the time taken for the arrow to hit the ground is approximately 0.639 seconds.
b) The horizontal distance traveled by the arrow can be found by considering the horizontal velocity and the time taken.
Given that the horizontal velocity (Vx) remains constant throughout the motion, we can use the formula:
Distance = velocity × time
The horizontal velocity of the arrow is given as 30 m/s, and the time taken is approximately 0.639 seconds:
Distance = 30 × 0.639
Distance ≈ 19.17 meters
Therefore, the horizontal distance traveled by the arrow is approximately 19.17 meters.
c) During the flight of the arrow:
i) The horizontal component of the velocity remains constant throughout the motion. Since there is no horizontal acceleration acting on the arrow (assuming no air resistance), the arrow will maintain the same horizontal velocity of 30 m/s throughout its flight.
ii) The vertical component of the velocity changes due to the acceleration of gravity. As the arrow moves upward, its vertical velocity decreases until it reaches its highest point, where it momentarily becomes zero. Then, as the arrow falls back down, the vertical velocity increases in magnitude at a rate of 9.8 m/s^2 due to the acceleration due to gravity.
a) To solve this problem, we can use the equations of motion for projectile motion. Since the arrow is fired horizontally, its initial velocity in the vertical direction is zero. We'll use the following equation to determine the time taken for the arrow to hit the ground:
h = ut + (1/2)gt^2
where h is the initial height (2m), u is the initial vertical velocity (0 m/s), g is the acceleration due to gravity (-9.8 m/s^2), and t is the time taken.
Substituting the given values into the equation:
2 = 0 + (1/2)(-9.8)t^2
2 = -4.9t^2
Dividing both sides by -4.9:
t^2 = 2/-4.9
Taking the square root of both sides:
t ≈ √(2/-4.9)
Using a calculator to find the square root and rounding to three significant figures:
t ≈ √(2/-4.9) ≈ 0.639 seconds (correct to 3 significant figures)
So, the time taken for the arrow to hit the ground is approximately 0.639 seconds.
b) Since the arrow was fired horizontally, the horizontal velocity remains constant throughout its flight. We can find the horizontal distance traveled by the arrow using the formula:
d = v * t
where d is the horizontal distance, v is the horizontal velocity, and t is the time taken (0.639 seconds).
The horizontal velocity (v) is given as 30 m/s, as the arrow is fired horizontally.
Plugging in the values:
d = 30 * 0.639
d ≈ 19.17 meters
So, the horizontal distance traveled by the arrow is approximately 19.17 meters.
c) During the flight of the arrow:
i) The horizontal component of the velocity remains constant at 30 m/s since there is no horizontal acceleration. This means that the magnitude of the horizontal component of the velocity remains unchanged.
ii) The vertical component of the velocity changes due to the acceleration due to gravity. As the arrow moves upward, the vertical component of the velocity decreases until it reaches its maximum height, at which point the vertical component becomes zero. As the arrow comes back down, the vertical component of the velocity increases in magnitude until it hits the ground.
Sure! I'll explain step by step how to solve each part of the problem.
a) To find the time taken for the arrow to hit the ground, we can use the formula for vertical motion. The vertical motion can be modeled using the equation:
h = ut + (1/2)gt^2
where:
- h is the height of the object (2m in this case),
- u is the initial vertical velocity (0 m/s as the arrow is fired horizontally),
- g is the acceleration due to gravity (approximately 9.8 m/s^2),
- t is the time taken for the arrow to hit the ground (what we're trying to find).
Since the initial vertical velocity, u, is 0 m/s, the equation simplifies to:
h = (1/2)gt^2
Rearrange the equation to solve for t:
t^2 = (2h) / g
t = sqrt((2h) / g)
Substitute the values given in the problem:
t = sqrt((2 * 2) / 9.8)
t = sqrt(4 / 9.8)
t ≈ 0.632 seconds
Therefore, the time taken for the arrow to hit the ground is approximately 0.632 seconds. Rounded to three significant figures, it becomes 0.639 seconds.
b) To find the horizontal distance traveled by the arrow, we can use the equation for horizontal motion. The horizontal motion can be modeled as a uniform motion, where the horizontal distance traveled is given by:
distance = speed * time
In this case, the speed of the arrow is 30 m/s and we have already found the time in part (a) to be 0.639 seconds:
distance = 30 m/s * 0.639 s
distance ≈ 19.17 meters
Therefore, the horizontal distance traveled by the arrow is approximately 19.17 meters.
c) i) During the flight of the arrow, the horizontal component of the velocity remains constant. Since there is no horizontal acceleration acting on the arrow, its horizontal velocity stays the same at 30 m/s throughout its flight.
c) ii) In the absence of air resistance, the only force acting on the arrow vertically is gravity. Gravity causes the vertical component of the velocity to change continuously. Initially, the vertical component is zero, but it increases downward due to the acceleration caused by gravity. Ultimately, when the arrow hits the ground, the vertical component of the velocity will be at its maximum, downward velocity.
I hope this explanation helps! Let me know if you have any further questions.