A projectile is fired with an initial speed of 31.0 at an angle of 10.0 above the horizontal. The object hits the ground 7.50 later.
a. How much higher or lower is the launch point relative to the point where the projectile hits the ground?
b. To what maximum height above the launch point does the projectile rise?
c. What is the magnitude of the projectile's velocity at the instant it hits the ground?
d. What is the direction (below +x) of the projectile's velocity at the instant it hits the ground?
With projectiles fired at an angle, we consider the vertical component (vy=v0 sin(θ)) and horizontal component (vx=v0 cos(θ)) separately.
vx stays constant throughout.
vy can be treated like a free fall after an iniital velocity of vy.
(a)
So assuming the projectile is fired at y=0, then calculate time t to reach maximum height attained using
t=vy/g
Subtract from total time T of 7.5 seconds to get the time of descent t1.
The distance descended, S1 is given
by
S1 = 0*t1+(1/2)(-g)t1²
Take the difference of S and S1 to get the difference of elevations. Watch the signs.
(b)
use S=(0-vy²)/(2(-g))
(c)
The vertical (downwards) velocity at impact is given by
vy1=t1*(-g)
Add this vectorially to vx to get the velocity.
(d)
the angle θ below the horizontal can be obtained by the arctangent of the ratio of vertical and horizontal components of velocity, namely:
tan(θ) = vy1/vx
(watch the signs, &theta should be between 0 and 90 degrees).
a. Well, it seems like the launch point is quite a "high" achiever because it managed to hit the ground after 7.50 seconds! So, the launch point is definitely "lower" than the point where the projectile hits the ground.
b. Ah, the projectile wants to reach for the stars! To find out how high it goes, we'll need to calculate the maximum height. But don't worry, it won't go so high that it needs a passport. Just a math problem. So, let's crunch those numbers!
c. To find the magnitude of the velocity at the instant it hits the ground, we'll need to use some physics magic. And by physics magic, I mean some good ol' mathematical calculations. Trust me, this will "velocity" your brain into action!
d. Ah, the direction of the velocity at the instant it hits the ground. Let's think about this... The projectile is hitting the ground, so it's going downwards. That means the direction is "below +x". Just like a graceful skydiver, this projectile is diving straight for the ground. Good thing it's not a clown projectile! They'd probably be doing somersaults on the way down. Keep it simple, projectile!
To solve these questions, we can use the equations of motion for projectile motion.
Let's break down the given information:
Initial speed (u) = 31.0 m/s
Launch angle (θ) = 10.0 degrees
Time of flight (t) = 7.50 s
a. To determine the vertical displacement, we can use the equation:
Δy = u * sin(θ) * t
Δy = 31.0 * sin(10.0) * 7.50
Δy ≈ 39.5 m
Therefore, the launch point is 39.5 m higher than the point where the projectile hits the ground.
b. To find the maximum height (h_max) reached by the projectile, we can use the equation:
h_max = (u^2 * (sin(θ))^2) / (2 * g)
h_max = (31.0^2 * (sin(10.0))^2) / (2 * 9.8)
h_max ≈ 4.01 m
The projectile reaches a maximum height of approximately 4.01 m above the launch point.
c. To calculate the magnitude of the projectile's velocity at the instant it hits the ground, we can use the horizontal component of the velocity since the vertical component is zero:
v_x = u * cos(θ)
v_x = 31.0 * cos(10.0)
v_x ≈ 30.17 m/s
Therefore, the magnitude of the projectile's velocity at the instant it hits the ground is approximately 30.17 m/s.
d. The direction of the projectile's velocity at the instant it hits the ground can be found using the angle between the x-axis and the velocity vector. Since the launch angle was given as 10.0 degrees above the horizontal, the velocity at the instant it hits the ground will be 10.0 degrees below the positive x-axis.
Therefore, the direction of the projectile's velocity at the instant it hits the ground is 10.0 degrees below +x.
To solve this problem, we can use the equations of motion for projectile motion. These equations describe the motion of an object in two dimensions under the influence of gravity. Here's how to approach each part of the problem:
a. To determine how much higher or lower the launch point is relative to the point where the projectile hits the ground, we need to calculate the vertical displacement. We can use the equation:
Δy = v₀y * t + (1/2) * g * t²
where Δy is the vertical displacement, v₀y is the initial vertical velocity, t is the time of flight, and g is the acceleration due to gravity.
Given that the initial speed is 31.0 m/s and the launch angle is 10.0°, we can find the initial vertical velocity:
v₀y = v₀ * sin(θ)
where θ is the launch angle in radians. Substituting the values:
v₀y = 31.0 * sin(10.0°)
With the known value for time of flight (t = 7.50 s), we can use the equation to calculate the vertical displacement. Since the projectile hits the ground, the vertical displacement will be negative.
Δy = v₀y * t + (1/2) * g * t²
b. To determine the maximum height above the launch point, we need to calculate the vertical displacement at the highest point of the trajectory. At the highest point, the vertical velocity will be zero. We can use the equation:
v_fy = v₀y + g * t_max
where v_fy is the final vertical velocity, v₀y is the initial vertical velocity, g is the acceleration due to gravity, and t_max is the time taken to reach the maximum height.
c. To find the magnitude of the projectile's velocity when it hits the ground, we can use the equation:
v_f = sqrt(v_fx² + v_fy²)
where v_f is the final velocity, v_fx is the final horizontal velocity, and v_fy is the final vertical velocity.
d. To find the direction (below +x) of the projectile's velocity at the instant it hits the ground, we need to calculate the angle of the velocity vector. We can use the equation:
θ = atan(v_fy / v_fx)
where θ is the angle of the velocity vector and atan is the inverse tangent function.
By following these steps, you should be able to find the answers to all parts of the problem.