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marusya05 [52]
3 years ago
8

A space vehicle deploys its re–entry parachute when it's traveling at a vertical velocity of –150 meters/second (negative becaus

e the parachute is going down.) It comes to rest at 0 meters/second after 30 seconds. What's the average acceleration of the shuttle during this time span?
Physics
1 answer:
dexar [7]3 years ago
6 0

Answer:

a=5m/s^2

Explanation:

Aceleration is a change on the velocity of the object in a given time.

For this case: the initial velocity is

v_{1}=150m/s

and the final velocity is :

v_{2}=0 m/s

so, the change in velocity is:

\Delta v =v_{2}-v_{1}=0m/s - (-150m/s) =  150 m/s

and the change in time , according to the problem:

\Delta t=30s

So, the aceleration is:

a=\frac{\Delta v}{\Delta t} = \frac{150m/s}{30s} = 5m/s^2

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Now she is out on a hike and comes to the left bank of a river. There is no bridge and the right bank is 10.0 mm away horizontal
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a. 8.96 m/s b. 1.81 m

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Here is the complete question.

a) A long jumper leaves the ground at 45° above the horizontal and lands 8.2 m away.

What is her "takeoff" speed  v 0 ?

b) Now she is out on a hike and comes to the left bank of a river. There is no bridge and the right bank is 10.0 m away horizontally and 2.5 m, vertically below.  

If she long jumps from the edge of the left bank at 45° with the speed calculated in part a), how long, or short, of the opposite bank will she land?

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b. We use R = v₀²sin2θ/g to calculate how long or short of the opposite bank she will land. With v₀ = 8.96 m/s and θ = 45

R = 8.96²sin(2 × 45)/9.8 = 80.2816/9.8 = 8.192 m.

So she land 8.192 m away from her bank. The distance away from the opposite bank she lands is 10 - 8.192 m = 1.808 m ≅ 1.81 m

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3 years ago
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