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dezoksy [38]
3 years ago
12

Help me answer this please!

Physics
1 answer:
Westkost [7]3 years ago
3 0

Answer:

Option iv

Explanation:

The acceleration due to gravity (g) of the earth is constant (i.e 9.81 ms¯²). As long as the stone remains within the earth, the acceleration due to gravity experienced by the stone will be 9.81 ms¯².

The maximum height reached by an object is the height where the object no longer have velocity to move further hence it begins to fall back. At this point (i.e at the maximum height), the velocity of the object is zero.

Summary:

At maximum height:

Acceleration due to gravity of the stone = 9.81 ms¯²

Velocity = 0

Thus, option iv gives the correct answer.

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A tiger leaps horizontally out of a tree that is 6.00 m high. If he lands 2.00 m from the base of the tree, calculate his initia
Musya8 [376]

Answer:

The initial speed of the tiger is 1.80 m/s

Explanation:

Hi there!

The equation of the position vector of the tiger is the following:

r = (x0 + v0 · t, y0 + 1/2 · g · t²)

Where:

r = position vector at a time t.

x0 = initial horizontal position.

v0 = initial horizontal velocity.

t = time.

y0 = initial vertical position,

g = acceleration due to gravity.

Let´s place the origin of the frame of reference on the ground at the point where the tree is located so that the initial position vector will be:

r0 = (0.00, 6.00) m

We can use the equation of the vertical component of the position vector to obtain the time it takes the tiger to reach the ground.

y = y0 + 1/2 · g · t²

When the tiger reaches the ground, y = 0:

0 = 6.00 m - 1/2 · 9.81 m/s² · t²

2 · (-6.00 m) / -9.81 m/s² = t²

t = 1.11 s

We know that in 1.11 s the tiger travels 2.00 m in the horizontal direction. Then, using the equation of the horizontal component of the position vector we can find the initial speed:

x = x0 + v0 · t

At t = 1.11 s, x = 2.00 m

x0 = 0

2.00 m = v0 · 1.11 s

2.00 m / 1.11 s = v0

v0 = 1.80 m/s

The initial speed of the tiger is 1.80 m/s

4 0
3 years ago
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