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Varvara68 [4.7K]
3 years ago
8

A student stands on the edge of a cliff that is 300 m high and kicks a rock horizontally. 7.8 seconds later, the rock hits the g

round. If the rock lands 50 meters from the base of the cliff, what was the rock's horizontal velocity?
Physics
1 answer:
mart [117]3 years ago
6 0

Answer:

6.65m/s

Explanation:

Using the equation of motion

S = ut + 1/2gt²

S is the height of fall

t is the time

u is the horizontal velocity

g is the acceleration due to gravity

Given

S = 300 + 50

S = 350m

t = 7.8seconds

g = 9.8m/s^2

Get S

S = 7.8u + 1/2(9.8)(7.8)²

S = 7.8u + 298.116

350 = 7.8u + 298.116

7.8u = 350 - 298.116

7.8u = 51.884

u = 51.884/7.8

u = 6.65m/s

Hence the rock's horizontal velocity was 6.65m/s

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The question is incomplete. Here is the complete question.

Cars A nad B are racing each other along the same straight road in the following manner: Car A has a head start and is a distance D_{A} beyond the starting line at t = 0. The starting line is at x = 0. Car A travels at a constant speed v_{A}. Car B starts at the starting line but has a better engine than Car A and thus Car B travels at a constant speed v_{B}, which is greater than v_{A}.

Part A: How long after Car B started the race will Car B catch up with Car A? Express the time in terms of given quantities.

Part B: How far from Car B's starting line will the cars be when Car B passes Car A? Express your answer in terms of known quantities.

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Explanation: First, let's write an equation of motion for each car.

Both cars travels with constant speed. So, they are an uniform rectilinear motion and its position equation is of the form:

x=x_{0}+vt

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x_{0} is initial position

v is velocity

t is time

Car A started the race at a distance. So at t = 0, initial position is D_{A}.

The equation will be:

x_{A}=D_{A}+v_{A}t

Car B started at the starting line. So, its equation is

x_{B}=v_{B}t

Part A: When they meet, both car are at "the same position":

D_{A}+v_{A}t=v_{B}t

v_{B}t-v_{A}t=D_{A}

t(v_{B}-v_{A})=D_{A}

t=\frac{D_{A}}{v_{B}-v_{A}}

Car B meet with Car A after t=\frac{D_{A}}{v_{B}-v_{A}} units of time.

Part B: With the meeting time, we can determine the position they will be:

x_{B}=v_{B}(\frac{D_{A}}{v_{B}-v_{A}} )

x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}

Since Car B started at the starting line, the distance Car B will be when it passes Car A is x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}} units of distance.

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