An arrow is fired with initial velocity v0 at an angle θ from the top of battlements, a height h above the ground. (Assume θ is measured above the horizontal.)
(a) In terms of h, v0, θ, and g, what is the time at which the arrow reaches its maximum height?
t =
(b) In terms of h, v0, θ, and g, what is the maximum height above the ground reached by the arrow?
hmax =
at the top, vertical velocity is zero.
vf=0=vo*sinTheta -gt
(vo*sinTheta-gt)=0
at the top, t=vo*sinTheta/g
At the top, height hmax
hmax=vo*sinTheta*t-4.8t^2
put the expression for t above into that to find hmax
(a) Well, it's time to give that arrow a taste of the high life! The time at which the arrow reaches its maximum height can be calculated using the following equation:
t = (v0 * sin(θ)) / g
(b) Ah, the elusive maximum height! Prepare to be amazed by this calculation:
hmax = (v0^2 * sin^2(θ)) / (2 * g) + h
Now, it's time to reach for the stars, or at least the maximum height!
(a) To find the time at which the arrow reaches its maximum height, we can use the equation for the vertical component of the arrow's velocity:
v_y = v0 * sin(θ) - g * t
At the maximum height, the vertical component of the velocity is zero, so we set v_y = 0 and solve for t:
0 = v0 * sin(θ) - g * t
Rearranging the equation:
g * t = v0 * sin(θ)
Dividing both sides by g:
t = v0 * sin(θ) / g
So, the time at which the arrow reaches its maximum height is t = v0 * sin(θ) / g.
(b) To find the maximum height above the ground reached by the arrow, we can use the equation for the vertical position of the arrow:
y = h + v0 * sin(θ) * t - (1/2) * g * t^2
At the maximum height, the vertical position is equal to hmax, so we set y = hmax and solve for t:
hmax = h + v0 * sin(θ) * t - (1/2) * g * t^2
Rearranging the equation:
(1/2) * g * t^2 - v0 * sin(θ) * t + (h - hmax) = 0
This is a quadratic equation in t. We can solve it using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = (1/2) * g, b = -v0 * sin(θ), and c = h - hmax. Substituting these values into the formula:
t = (-(-v0 * sin(θ)) ± √((-v0 * sin(θ))^2 - 4 * (1/2) * g * (h - hmax))) / (2 * (1/2) * g)
Simplifying further:
t = (v0 * sin(θ) ± √(v0^2 * sin^2(θ) + 2g * (h - hmax))) / g
So, the equation for the maximum height above the ground reached by the arrow is hmax = h + v0 * sin(θ) * t - (1/2) * g * t^2.
To calculate the time at which the arrow reaches its maximum height (t), and the maximum height above the ground reached by the arrow (hmax), we can use the equations of motion for projectile motion.
(a) To find the time at which the arrow reaches its maximum height, we need to calculate the time it takes for the vertical component of the velocity to become zero. At its maximum height, the vertical velocity is zero.
The initial vertical velocity (Vy) can be calculated as: Vy = v0 * sin(θ)
Since the acceleration due to gravity (g) acts downwards, the vertical acceleration (Ay) is equal to -g.
Using the equation of motion: Vy = Vy0 + Ay * t
At maximum height, Vy = 0, and substituting the values we have:
0 = v0 * sin(θ) - g * t_max
Rearranging the equation, we get:
v0 * sin(θ) = g * t_max
So, t_max = (v0 * sin(θ)) / g
Therefore, the time at which the arrow reaches its maximum height is t = (v0 * sin(θ)) / g.
(b) To find the maximum height above the ground reached by the arrow (hmax), we can use the equation for vertical displacement:
h = h0 + Vy0 * t + (1/2) * Ay * t^2
At maximum height, the vertical displacement (h) is equal to hmax, the initial vertical position (h0) is zero, the initial vertical velocity (Vy0) is v0 * sin(θ), and the vertical acceleration (Ay) is -g.
Therefore, the equation becomes:
hmax = 0 + (v0 * sin(θ)) * t_max + (1/2) * (-g) * t_max^2
Substituting the equation for t_max from part (a), we get:
hmax = 0 + (v0 * sin(θ)) * [(v0 * sin(θ)) / g] + (1/2) * (-g) * [(v0 * sin(θ)) / g]^2
Simplifying the equation, we get:
hmax = (v0^2 * sin^2(θ)) / (2 * g)
Therefore, the maximum height above the ground reached by the arrow is hmax = (v0^2 * sin^2(θ)) / (2 * g).