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Naddik [55]
3 years ago
10

A boy throws a rock with an initial velocity of 2.15 m/s at 30.0° above the horizontal. If air resistance is negligible, how lon

g does it take for the rock to reach the maximum height of its trajectory?
Physics
1 answer:
vladimir1956 [14]3 years ago
4 0

Answer:

t = 0.11 second

The time taken to reach the maximum height of the trajectory is 0.11 second

Explanation:

Taking the vertical component of the initial velocity;

Vy = Vsin∅

Initial velocity V = 2.15 m/s

Angle ∅ = 30°

Vy = 2.15sin30 = 2.15 × 0.5

Vy = 1.075 m/s

The height of the rock at time t during the flight is;

From the equation of motion;

h(t) = Vy×t - 0.5gt^2

g= acceleration due to gravity = 9.8m/s^2

Substituting the given values;

h(t) = 1.075t - 0.5(9.8)t^2

h(t) = 1.075t - 4.9t^2

The rock is at maximum height when dh/dt = 0;

dh(t)/dt = 1.075 - 9.8t = 0

1.075 - 9.8t = 0

9.8t = 1.075

t = 1.075/9.8

t = 0.109693877551 s

t = 0.11 second

The time taken to reach the maximum height of the trajectory is 0.11 second

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We drive at a speed of 20 km/h for 3 hours. Then we drive 4 hours at 30 km/h. Calculate our average speed.
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4 0
2 years ago
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2 years ago
4. How long will it take a car travelling with a speed of 160 km hr to cover a distance of 700 meters? Hint: km/hr should be con
Inessa [10]

Answer:

15.8 seconds

Explanation:

Create an extended calculation to convert all the unit to what you need.

160 km      1000 m       1 hour         1 min

----------- x ------------- x -------------- x ----------   =  44.4 m/s

1 hour            1 km         60 min      60 sec

So 160km/hr is equal to 44.4m/s

Now you can figure out how many seconds it will take to go 700 meters.

44.4 m          

----------   X     x sec   =  700 m

1  sec

Solve for x sec

x sec = 700m / 44.4 m/s

         =  15.8 seconds

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2 years ago
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