Which parameter of a projectile depends on the horizontal as well as the vertical component of velocity of projection?

Physics

2 answers:

Maurinko [17]2 years ago

7 0

Answer:

It RANGE

Explanation:

I just did it and that's the right answer. Angle of launch creates the horizontal and vertical velocity components, it is not dependent on them!!!!

jekas [21]2 years ago

6 0

Just about anything you could ask about a projectile AFTER it's launched
depends on both components of the launch velocity.

Here are some that I can think of:

-- angle of launch
-- magnitude of launch velocity
-- location at any time after launch
-- magnitude of velocity at any time after launch
-- direction of velocity at any time after launch
-- distance of the landing point from the launch point

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2 years ago

Consider a uniform horizontal wooden board that acts as a pedestrian bridge. The bridge has a mass of 300 kg and a length of 10

gayaneshka [121]

Answer:

F = 2123.33N

Explanation:

In order to calculate the torque applied by the left support, you take into account that the system is at equilibrium. Then, the resultant of the implied torques are zero.

$\Sigma \tau=0$

Next, you calculate the resultant of the torques around the right support, by taking into account that the torques are generated by the center of mass of the wooden, the person and the left support. Furthermore, you take into account that torques in a clockwise direction are negative and in counterclockwise are positive.

Then, you obtain the following formula:

$-\tau_l+\tau_p+\tau_{cm}=0$ (1)

τl: torque produced by the left support

τp: torque produced by the person

τcm: torque produced by the center of mass of the wooden

The torque is given by:

$\tau=Fd$ (2)

F: force applied

d: distance to the pivot of the torque, in this case, distance to the right support.

You replace the equation (2) into the equation (1) and take into account that the force applied by the person and the center of mass of the wood are the their weight:

$-Fd_1+W_pd_2+W_{cm}d_3=0\\\\d_1=6.0m\\\\d_2=2.0m\\\\d_3=3.0m\\\\W_p=(200kg)(9.8m/s^2)=1960N\\\\W_{cm}=(300kg)(9.8m/s^2)=2940N$