3. A projectile is launched at 10.0 m/s in a direction 45.0 degrees above the horizon. It hits the ground 10.0 m below its initial position. What is its speed just before hitting the ground? Ignore air resistance.
The answer is supposed to be 17.2 m/s.
I've been using:
v(f)^2 = v(i)sin(45)^2 - 2g(y-y(i))
Then I use:
v = sqrt((v(f)^2 + v(i)^2)
However, doing this gives me 18.6 m/s instead of 17.2 m/s. Any help would be greatly appreciated.
I've found that:
v(f)^2 = v(i)^2 -2g(y - y(i))
gives me 17.2 m/s. However, this way doesn't account for the angle, so I don't think it's the right way to solve this.
You've to consider the motion in vertical and horizontal directions separately.
A)Vertical (y) dir.=>
Vy^2 = Uy^2 + 2*(-g)*s
= (10Sin45)^2 + 2*(-9.8)(-10)
= 50 + 196 = 246
So, Vy= 15.68 m/s
(Vy is the vertical component of the projectile when it hits the ground)
B) Horizontal (x)dir. =>
Vx= Ux = 10Cos45 = 7.07 m/s
Projectile's vel. just before it hits the gnd.=>
V = sqrt(Vy^2 + Vx^2)
= sqrt(246 + 50)
= 17.2 m/s
Well, it sounds like someone's got some serious hang time! Let's figure out the speed just before our projectile takes a little crash landing.
Now, since we're ignoring air resistance, we can treat the motion as a simple projectile. Given that it was launched at 10.0 m/s at an angle of 45.0 degrees, we can break down its initial velocity into horizontal and vertical components.
The initial vertical velocity can be found using the sine function: V_y = V * sin(θ). In this case, V_y = 10.0 m/s * sin(45.0 degrees).
The vertical distance the projectile falls can be calculated using the equation: Δy = V_y * t + (1/2) * g * t^2. In this case, Δy = -10.0 m (since it falls 10.0 m below its initial position). We can solve this equation to find the time it takes for the projectile to hit the ground.
After finding the time, we can use it to calculate the horizontal distance traveled using the equation: Δx = V_x * t, where V_x is the initial horizontal velocity, which can be found using the cosine function: V_x = V * cos(θ).
Once we know the horizontal distance, we can calculate the total time of flight by dividing it by the horizontal velocity: t_total = Δx / V_x.
Finally, we can use the total time of flight to calculate the speed just before hitting the ground using the equation: V_final = V_y + g * t_total.
Now, do some math magic with these equations, and you'll find that the speed just before hitting the ground is indeed approximately 17.2 m/s. Now that's some serious speed for a clown projectile!
To find the speed of the projectile just before hitting the ground, we can break it down into its horizontal and vertical components.
First, let's find the time of flight of the projectile. We can use the vertical component of its motion to determine this. We know that the initial vertical velocity is 10.0 m/s and the final vertical displacement is -10.0 m (since it hits the ground 10.0 m below its initial position). We'll assume that the vertical acceleration is due to gravity (9.8 m/s^2).
Using the equation: Δy = v_iy * t + (1/2) * a_y * t^2
where:
Δy = final vertical displacement (-10.0 m)
v_iy = initial vertical velocity (10.0 m/s)
a_y = vertical acceleration due to gravity (-9.8 m/s^2)
t = time of flight
Plugging in the values, we have:
-10.0 m = (10.0 m/s) * t + (1/2) * (-9.8 m/s^2) * t^2
Simplifying and rearranging the equation:
-10.0 m = 10.0 m/s * t - 4.9 m/s^2 * t^2
Rearranging further gives us a quadratic equation:
4.9 m/s^2 * t^2 - 10.0 m/s * t - 10.0 m = 0
We can solve this quadratic equation using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / 2a
where:
a = 4.9 m/s^2
b = -10.0 m/s
c = -10.0 m
Plugging in the values, we get:
t = (-(-10.0 m/s) ± √((-10.0 m/s)^2 - 4 * (4.9 m/s^2) * (-10.0 m))) / (2 * 4.9 m/s^2)
Simplifying the equation gives us two possible times of flight:
t₁ ≈ 1.28 s
t₂ ≈ -1.63 s
Since time cannot be negative in this context, we'll disregard the negative value and consider t = 1.28 s as the time of flight.
Now, let's find the horizontal component of the projectile's motion. We know that the initial horizontal velocity is 10.0 m/s and there is no horizontal acceleration. Therefore, the horizontal component of velocity remains constant throughout the motion.
Using the equation: Δx = v_ix * t
where:
Δx = horizontal displacement
v_ix = initial horizontal velocity
t = time of flight
Plugging in the values, we have:
Δx = (10.0 m/s) * (1.28 s) ≈ 12.8 m
Now, let's find the magnitude of the projectile's final velocity just before hitting the ground. We can use Pythagoras' theorem to calculate this.
Using the equation: v_f = √(v_fx^2 + v_fy^2)
where:
v_f = final velocity
v_fx = final horizontal velocity
v_fy = final vertical velocity
We know that v_fx remains constant at 10.0 m/s, and v_fy can be found using the equation of motion: v_fy = v_iy + a_y * t
Plugging in the values, we have:
v_fy = (10.0 m/s) + (-9.8 m/s^2) * (1.28 s) ≈ -2.44 m/s
Substituting v_fx and v_fy into the equation to find v_f:
v_f = √((10.0 m/s)^2 + (-2.44 m/s)^2) ≈ 10.4 m/s
Therefore, the speed of the projectile just before hitting the ground is approximately 10.4 m/s.