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chubhunter [2.5K]
2 years ago
7

A ball is released from the top of a tower of height h meters. it takes T seconds to reach the ground . what is the position of

the ball in T/3 seconds ?​
Physics
1 answer:
alekssr [168]2 years ago
3 0

Answer:the

8/9 h

Explanation:

Height  =   1/2 a T^2     now change to T/3

 now height = 1/2 a (T/3)^2   =<u> 1/9</u>  1/2 a T^2     <===== it is 1/9 of the way down   or   8/9 h

 

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8 0
4 years ago
a water bomber flying with a horizontal speed of 85m/s at a height of 3000m drops a load on a fire below. How far in front of th
Andreyy89

Answer:

2081.65 m

Explanation:

We'll begin by calculating the time taken for the load to get to the target. This can be obtained as follow:

Height (h) = 3000 m

Acceleration due to gravity (g) = 10 m/s²

Time (t) =?

h = ½gt²

3000 = ½ × 10 × t²

3000 = 5 × t²

Divide both side by 5

t² = 3000 / 5

t² = 600

Take the square root of both side

t = √600

t = 24.49 s

Finally, we shall determine the distance from the target at which the load should be released. This can be obtained as follow:

Horizontal velocity (u) = 85 m/s

Time (t) = 24.49 s

Horizontal distance (s) =?

s = ut

s = 85 × 24.49

s = 2081.65 m

Thus, the load should be released from 2081.65 m.

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3 years ago
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5 0
3 years ago
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Hola tengo un taller de física y nose como resolver este pregunta
Blizzard [7]

a) 0.26 h

b) 71.4 km

Explanation:

a)

In order to solve the problem, we have to know what is the final velocity of the car.

Here, we assume that the final velocity reached by the car is

v=300 km/h

Therefore, we can find the time taken by the car to reach this velocity by using the suvat equation:

v=u+at

where:

u = 250 km/h is the initial velocity

a=190 km/h^2 is the acceleration of the car

v = 300 km/h is the final velocity

t is the time

Solving for t, we find:

t=\frac{v-u}{a}=\frac{300-250}{190}=0.26 h

b)

In order to find the distance covered by the car, we can use the following suvat equation:

s=ut+\frac{1}{2}at^2

where:

s is the distance covered

u is the initial velocity

a is the acceleration

t is the time

For the car in this problem, we have:

u = 250 km/h

t = 0.26 h (calculated in part a)

a=190 km/h^2

Therefore, the distance covered is

s=(250)(0.26)+\frac{1}{2}(190)(0.26)^2=71.4 km

8 0
3 years ago
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