Rock A is thrown horizontally off of a cliff with a velocity of 20m/s. The rock lands 60m from the base of the cliff.

Physics

1 answer:

mixas84 [53]2 years ago

4 0

Answer:

B. 44.1

Explanation:

We can use the equation y=1/2gt^2 to find the distance in the vertical direction, where y= distance in the vertical direction, g=acceleration due to gravity, and t=time

we only know acceleration due to gravity(9.8m/s^2),

plug in known variables,

y=1/2*9.8*t^2

we still need to plug in the time variable.

In order to do that, we an use the s=d/t equation and re arrange that formula to get: t=d/s, where t=time, d=distance, and s=speed.

plug in known variables,

t=60/20

t=3

now that we know the time, we can plug that into our original equation

y=1/2*9.8*3^2

simplify,

y=4.9*3^2

y=4.9*9

y=44.1m

Thus, B is your answer

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The electric field in a region is uniform (constant in space) and given by E-( 148.0 1 -110.03)N/C. An additional charge 10.4 nC

enyata [817]

Answer:

The y-component of the electric force on this charge is $F_y = -1.144\times 10^{-6}\ N.$

Explanation:

<u>Given:</u>

Electric field in the region, $\vec E = (148.0\ \hat i-110.0\ \hat j)\ N/C.$
Charge placed into the region, $q = 10.4\ nC = 10.4\times 10^{-9}\ C.$

where, $\hat i,\ \hat j$ are the unit vectors along the positive x and y axes respectively.

The electric field at a point is defined as the electrostatic force experienced per unit positive test charge, placed at that point, such that,

$\vec E = \dfrac{\vec F}{q}\\\therefore \vec F = q\vec E\\=(10.4\times 10^{-9})\times (148.0\ \hat i-110.0\ \hat j)\\=(1.539\times 10^{-6}\ \hat i-1.144\times 10^{-6}\ \hat j)\ N.$