A girl throws a marshmallow that lands in her friend’s mouth 2 m away. The girl threw the
marshmallow at an angle of 30 degrees. How hard did she throw the marshmallow?
To determine how hard the marshmallow was thrown, we can use the principles of projectile motion.
Step 1: Break the initial velocity into horizontal and vertical components.
The horizontal component (Vx) remains constant throughout, while the vertical component (Vy) changes due to gravity.
The initial velocity (V) can be calculated using the given angle and the total distance covered by the marshmallow (2 m):
V = 2 m / cos(30 degrees)
Step 2: Calculate the vertical component (Vy) of the initial velocity.
The vertical velocity can be determined using the sine of the angle (30 degrees) and the magnitude of the initial velocity (V):
Vy = V * sin(30 degrees)
Step 3: Calculate the horizontal component (Vx) of the initial velocity.
The horizontal velocity remains constant and can be determined using the cosine of the angle (30 degrees) and the magnitude of the initial velocity (V):
Vx = V * cos(30 degrees)
Step 4: The resulting initial velocity can be determined using the Pythagorean theorem:
V^2 = Vx^2 + Vy^2
By following these steps, you can calculate the initial velocity at which the girl threw the marshmallow.
To determine how hard the girl threw the marshmallow, we need to find its initial velocity. Since we know the distance the marshmallow traveled (2 m) and the launch angle (30 degrees), we can use projectile motion equations to find the initial velocity.
First, we need to split the initial velocity into horizontal (Vx) and vertical (Vy) components. The horizontal component (Vx) remains constant throughout the motion and is given by Vx = V0 * cos(theta), where V0 is the initial velocity and theta is the launch angle.
In this case, Vx = V0 * cos(30 degrees).
Next, we can determine the vertical component (Vy) using the equation Vy = V0 * sin(theta).
Knowing that the horizontal displacement (2 m) is related to the horizontal velocity (Vx) and the time of flight (t) by the equation s = Vx * t, we can rearrange the equation to find t: t = s / Vx.
Since the time of flight for the projectile motion is the same for the vertical component, we can determine the vertical displacement (h) using the equation h = Vy * t + 0.5 * g * t^2, where g is the acceleration due to gravity.
In this case, the vertical displacement (h) is 0 because the marshmallow lands at the same height from where it was launched.
Combining the equations, we can now solve for the initial velocity (V0): h = Vy * (s / Vx) + 0.5 * g * (s / Vx)^2.
Since h = 0, the equation simplifies to 0 = Vy * (s / Vx) + 0.5 * g * (s / Vx)^2.
Substituting the known values: 0 = V0 * sin(30 degrees) * (2 / V0 * cos(30 degrees)) + 0.5 * g * (2 / V0 * cos(30 degrees))^2.
Simplifying further: 0 = sin(30 degrees) * (2 / cos(30 degrees)) + 0.5 * g * (2 / cos(30 degrees))^2.
Using the values of sin(30 degrees) = 0.5 and cos(30 degrees) = sqrt(3) / 2, we can calculate: 0 = 0.5 * (2 / (sqrt(3) / 2)) + 0.5 * g * (2 / (sqrt(3) / 2))^2.
Now we can solve for g, the acceleration due to gravity, which is approximately 9.8 m/s^2.
Using this value in the equation, we can solve for V0.
well, I guess I assume the mouth and the throwing hand are at the same height, call it h = 0
S is the speed we want
u = S cos 30
Vi = S sin 30
v =
h = 0 + Vi t - 4.9 t^2
when is h = 0 (same height as at start)
4.9 t^2 -Vi t = 0
(4.9 t -Vi)t = 0
so t = 0 (that was at the start
or t = Vi/4.9
so
Vi = 4.9 t = S sin 30
t = S sin 30/4.9
now the horizontal problem
2 meters = u t
2 = S cos 30 * t
t = 2/(S cos 30) = S sin 30/4.9
9.8 = S^2 sin 30 cos 30
S^2 = 9.8 / (sin 30 * cos 30)