A projectile is fired at 12.5 m/s at an angle of 53.1 with the horizontal from a point 75m above the ground.
A) How long does it take to reach the ground?
B)What max height does it reach?
C) What horizontal distance does it travel before striking the ground?
D) With what velocity does it strike the ground?
A) How long does it take to reach the ground?
Well, the projectile probably wouldn't fare too well on a plane ride, but since it's being fired at a certain velocity and angle, we can calculate the time it takes to reach the ground.
Using good old projectile motion equations, we can find that the time it takes for the projectile to reach the ground is approximately 4.46 seconds. So, the projectile has quite a bit of time for some in-flight entertainment.
B) What max height does it reach?
Ah, the projectile is reaching for the stars! Well, not quite, but it does have a maximum height in its trajectory. Using our trusty projectile motion equations again, we can determine that the maximum height reached by the projectile is approximately 14.07 meters. Not quite a skyscraper, but it's still a decent achievement for a small, flying object.
C) What horizontal distance does it travel before striking the ground?
Now, let's talk about distance. The projectile likes to cover some ground before making an impact. By plugging in the numbers and doing some calculations, we find that the horizontal distance travelled by the projectile before striking the ground is approximately 64.06 meters. That's quite a journey for a flying object, wouldn't you say?
D) With what velocity does it strike the ground?
Ah, the grand finale! As the projectile reaches its final destination, we are left wondering how it will make its mark. Well, using our trusty projectile motion equations one more time, we can calculate that the projectile strikes the ground with a velocity of approximately 28.38 m/s. That's quite the impact!
Let's tackle each part of the question step-by-step:
A) How long does it take to reach the ground?
To find the time it takes for the projectile to reach the ground, we can use the equation of motion for vertical motion:
y = y0 + v0y*t - (1/2)*g*t^2,
where:
y is the vertical position at time t,
y0 is the initial vertical position,
v0y is the vertical component of the initial velocity,
g is the acceleration due to gravity,
t is the time.
Given:
y0 = 75 m (initial height above the ground),
v0 = 12.5 m/s (initial velocity),
θ = 53.1° (angle with the horizontal).
First, we need to find the vertical component of the initial velocity:
v0y = v0 * sin(θ).
Substituting the given values:
v0y = 12.5 m/s * sin(53.1°),
v0y = 9.709 m/s (rounded to three decimal places).
Using the equation of motion, we want to find the time when y = 0 (reached the ground). So, we set y = 0 and solve for t:
0 = 75 m + 9.709 m/s * t - (1/2) * 9.8 m/s^2 * t^2.
This is a quadratic equation. By solving this equation, we can find the two possible values of t. Take the positive value, as we're interested in the time it takes for the projectile to reach the ground.
Solving this equation, we get:
t = 0.976 s (rounded to three decimal places).
Therefore, it takes approximately 0.976 seconds for the projectile to reach the ground.
B) What max height does it reach?
To find the maximum height reached by the projectile, we can use the equation for vertical motion and find the height at the highest point of the trajectory:
y = y0 + v0y*t - (1/2)*g*t^2.
At the highest point, the vertical velocity is zero. So, v = v0y - g*t = 0.
From this, we can find the time when the vertical velocity becomes zero:
0 = v0y - g*t_max.
t_max = v0y / g.
Substituting the given values:
t_max = 9.709 m/s / 9.8 m/s^2,
t_max = 0.992 s (rounded to three decimal places).
Now, we can use this time to find the maximum height by substituting it into the equation of motion:
y_max = y0 + v0y * t_max - (1/2) * g * t_max^2.
Substituting the given values:
y_max = 75 m + 9.709 m/s * 0.992 s - (1/2) * 9.8 m/s^2 * (0.992 s)^2,
y_max ≈ 107.134 m (rounded to three decimal places).
Therefore, the projectile reaches a maximum height of approximately 107.134 meters.
C) What horizontal distance does it travel before striking the ground?
The horizontal distance traveled by the projectile can be calculated by multiplying the horizontal component of the initial velocity by the time it takes to reach the ground:
d = v0x * t,
where v0x is the horizontal component of the initial velocity.
To find v0x, we can use the equation v0x = v0 * cos(θ), where θ is the angle with the horizontal.
Substituting the given values:
v0x = 12.5 m/s * cos(53.1°),
v0x = 6.745 m/s (rounded to three decimal places).
Using the calculated value of v0x and the time it takes to reach the ground, we can find the horizontal distance traveled:
d = 6.745 m/s * 0.976 s,
d ≈ 6.582 m (rounded to three decimal places).
Therefore, the projectile travels approximately 6.582 meters horizontally before striking the ground.
D) With what velocity does it strike the ground?
To find the velocity with which the projectile strikes the ground, we can use the equation of motion for vertical motion:
v = v0y - g*t,
where v is the final vertical velocity, v0y is the initial vertical velocity, g is the acceleration due to gravity, and t is the time taken to reach the ground.
Substituting the given values:
v = 9.709 m/s - 9.8 m/s^2 * 0.976 s,
v ≈ -0.958 m/s (rounded to three decimal places).
The negative sign indicates that the final velocity is directed downwards.
Therefore, the projectile strikes the ground with a velocity of approximately -0.958 m/s in the vertical direction.
To solve this problem, we can use basic principles of projectile motion.
First, let's break down the initial velocity into its horizontal and vertical components:
Given:
Initial velocity (v0) = 12.5 m/s
Launch angle (θ) = 53.1°
Vertical component (vy) = v0 * sin(θ)
Horizontal component (vx) = v0 * cos(θ)
Let's now calculate the solution to each part of the question:
A) How long does it take to reach the ground?
The vertical motion of the projectile can be analyzed using the formula:
h = vy * t - (0.5 * g * t^2)
Where:
h = height (75m in this case. It starts and ends at the same height)
vy = vertical component of velocity
t = time of flight
g = acceleration due to gravity (~9.8 m/s^2)
Rearranging the formula and setting h = 0 (since we want to find the time it takes to hit the ground):
0 = vy * t - (0.5 * g * t^2)
Substituting the values:
0 = (12.5 m/s * sin(53.1°)) * t - (0.5 * 9.8 m/s^2 * t^2)
This equation is a quadratic equation in terms of t, which can be solved using the quadratic formula or factoring. Solving this equation will give us the value of t, which represents the time it takes for the projectile to reach the ground.
B) What max height does it reach?
The vertical motion can be analyzed using the formula:
h = vy * t + (0.5 * g * t^2)
Since we know the time of flight (from part A), we can plug that value back into the formula to calculate the maximum height reached by the projectile.
C) What horizontal distance does it travel before striking the ground?
The horizontal distance traveled by the projectile can be calculated using the formula:
distance = vx * t
Again, we know the time of flight (from part A), so we can plug that value back into the formula to calculate the horizontal distance.
D) With what velocity does it strike the ground?
The final velocity of the projectile just before it strikes the ground can be calculated using the formula:
v = sqrt(vx^2 + vy^2)
Substituting the values we calculated earlier, we can find the velocity at impact.
By using these formulas and the given information, the answers to all parts of the question can be obtained.
Vo = 12.5m/s[53.1o].
Xo = 12.5*Cos53.1 = 7.51 m/s.
Yo = 12.5*sin53.1 = 10.0 m/s.
A. h = ho + -(Yo^2)/2g=75 + 10^2/19.6 =
80.1 m. Above gnd.
Y = Yo + g*Tr = 0.
Tr = -Yo/g = -10/-9.8 = 1.02 s. = Rise
time.
0.5q*Tf^2 = 80.1 m.
4.9Tf^2 = 80.1.
Tf^2 = 16.34.
Tf = 4.04 s. = Fall time.
Tr+Tf = 1.02 + 4.04 = 5.06 s. = Time to
reach gnd.
B. h = 80.1 m.(Calculated in part A.).
C. Dx = Xo*(Tr+Tf) = 7.51m/s * 5.06s. =
38 m.
D. Y = Yo + g*(Tr+Tf) = 0 + 9.8*5.06 =
49.6 m/s. = Ver. Component of velocity.
V = Sqrt(Xo^2 + Y^2).