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3 years ago

How to find initial velocity in projectile motion without angle?

Physics

1 answer:

lisabon 2012 [21]3 years ago

6 0

Refer to the diagram shown below.

We want to find V, the initial launch velocity of a projectile, without knowing θ, the launch angle.

To find V, we need to perform an experiment to determine:
d, the horizontal distance traveled,
t, the time taken to travel the horizontal distance.

Assume that aerodynamic resistance is negligible.
The horizontal component of the velocity is
Vx = V cosθ
Because the horizontal distance, d, is traveled in time, t, therefore
(V cosθ)*t = d
Vcosθ = d/t (1)

Assume that ground level has height zero.
Note that g, the acceleration due to gravity is known.

The vertical travel between the time of launch and return to ground level obeys the equation
Vsinθ*t - (g/2)*t² = 0
Therefore
t[Vsinθ - (gt)/2] = 0
Obtain
t = 0, which corresponds to launch
or
t = (2Vsinθ)/g

That is,
Vsinθ = (gt)/2 (2)

From (1) and (2), obtain
(Vcosθ)² + (Vsinθ)² = (d/t)² + (g²t²)/4
V²(cos² + sin² ) = (d/t)² + (g²t²)/4
$V= \sqrt{( \frac{d}{t})^{2}+( \frac{gt}{2})^{2} }$

Because g,t and d are known, V can be calculated.

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A dockworker applies a constant horizontal force of 73.0 N to a block of ice on a smooth horizontal floor. The frictional force

vivado [14]

Answer:

a) 57.0 kg b) 24.2 m

Explanation:

a) According Newton's second law, the applied force is equal to the product of the mass times the acceleration.

As the force is constant, the acceleration is constant too.

In this case, as we have as givens the distance and the time, and also we know that the block is starting form rest, we can get the acceleration as follows:

d = 1/2 * a * t² ⇒ a = 2d / t² ⇒ a= 2* 13.0 m / (4.5)² s² = 1.28 m/s²

Replacing in the Newton's 2nd Law equation:

F = m*a ⇒ m = F/a = 73.0 N / 1.28 m/s = 57.0 Kg

b) At t=4.5 s, applying the definition of acceleration, we can get the value of the velocity at that time, as follows:

v= a* t = 1.28 m/s * 4.5 s = 5.76 m/s

If the worker stops pushing at the end of the 4. 5 s, this means (neglecting friction) that from that time omwards, no net force acts on the block, so it continues moving at constant speed.

In order to get the distance moved in the next 4.20 sec, as it is moving at constant speed, we neeed just to apply the definition of velocity:

v= Δx / Δt ⇒ Δx = v* Δt = 5.76 m/s * 4.2 m = 24.2 m

So, the total distance traveled during all the time (9.1 s) is just the sum of the 13.0 m advanced during the time when there was a constant force applied, and the last 24.2 m at constant speed, as follows:

d = 13.0 + 24.2 = 37.2 m

3 0

3 years ago

3 kg of saturated liquid water, contained in a rigid container at p = 1.5 bar, is heated by a 80 W heater. a) What will the phas

Mkey [24]

Is this science also? bc i know that

5 0

3 years ago

At what angle do the cars move after the collision?

liubo4ka [24]

Answer:

his results in the final angle after the collision of 37.2 degrees basically what we did there is turn the vector into a right triangle. We use sohcahtoa to solve for the angle. Being.

Explanation:

6 0

2 years ago

A star is estimated to be 4.8 km away from the earth. When we see this star in the night sky how old is the image. Let the speed

andre [41]

Time = distance / speed

Time = (4,800 meters) / (3 x 10⁸ m/s)

Time = 0.000016 second

This number is not one of the choices on the list. My hunch is that you copied the distance wrong.

If the estimated distance to the star is actually 4.8 x 10¹⁵ km, instead of 4.8 km, then the answer would be close to 500 years (B).

There's no way a star can be "4.8 km away from the Earth". You can walk that far in about an hour, and passenger jet airplanes fly twice as far as that away from the Earth !

5 0

3 years ago

The spring is unstretched at the position x = 0. under the action of a force p, the cart moves from the initial position x1 = -8

hammer [34]

Missing figure and missing details can be found here:
http://d2vlcm61l7u1fs.cloudfront.net/media%2Fdd5%2Fdd5b98eb-b147-41c4-b2c8-ab75a78baf37%2FphpEgdSbC....

Solution:
(a) The work done by the spring is given by
$W= \frac{1}{2} k (\Delta x)^2 
$
where k is the elastic constant of the spring and $\Delta x$ is the stretch between the initial and final position. Since x1=-8 in=-0.203 m and x2=5 in=0.127 m, we have
$W= \frac{1}{2} \cdot 500 N/m \cdot (0.127m-(-0.203m))^2=27.25 J$

(b) The work done by the weight is the product of the component of the weight parallel to the inclined plane and the displacement of the cart:
$W_W = -F_{//} (x_2 -x_1)$
where the negative sign is given by the fact that $F_{//}$ points in the opposite direction of the displacement of the cart, and where
$F_{//}=m g sin 15^{\circ}=6 kg \cdot 9.81m/s^2 \cdot sin 15^{\circ}=15.2 N$
therefore, the work done by the weight is
$W_W=-15.2 N \cdot (0.203m-(-0.127m))=-5.02 J$

8 0

3 years ago