A tiger leaps horizontally from a 12 m high rock with a speed of 4.5 m/s. how far from the base of the rock will she land?

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asifejaz20 asifejaz20 · Answer 1 · 2020-01-01T12:26:37+01:00

Answer:

7.02 m away from the base of the rock.

Explanation:

As the tiger leaps from 12 m high rock so, its initial velocity will be zero. According to second equation of motion:

h = (Vi)(t)+½ gt^2

12 = (0)(t)+1/2(9.8)t^2

t^2 = (12)(2)/(9.8)

t = √2.44

t = 1.56 s

Now,

X = (V)(t)

where,

V = average velocity = 4.5 m/s

X = (4.5)(1.56)

X = 7.02 m

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raphealnwobi raphealnwobi · Answer 2 · 2021-11-03T15:37:48+01:00

The distance from the base of the rock to the point the rock would land is 7 meters.

An object thrown into space experiences projectile motion.

The range of a projectile is the horizontal distance between the launch point and landing point.

Given g = 10 m/s², v = final velocity, height (h) of 12 m and initial velocity (u) of 4.5 m/s, the time taken to hit the ground is:

Δh = ut + 0.5gt²

-12 = 4.5sin(0) - 0.5(10)t²

5t² = 12

t = 1.55 seconds

Range = ucos(0) * t= 4.5 * cos(0) * 1.55 = 7 meters

The distance from the base of the rock to the point the rock would land is 7 meters.

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