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Bezzdna [24]
2 years ago
15

An object is thrown upward with some velocity. If the object rises 77.5 m above the point of release, (a) how fast was the objec

t thrown?, (b) How long did it take for the object to reach it's highest point ?
Physics
1 answer:
jolli1 [7]2 years ago
7 0

Answer:

v_o=39\ m/s\\t_m=4\ s

Explanation:

<u>Vertical Launch Upwards</u>

In a vertical launch upwards, an object is launched vertically up from a height H without taking into consideration any kind of friction with the air.

If vo is the initial speed and g is the acceleration of gravity, the maximum height reached by the object is given by:

\displaystyle h_m=H+\frac{v_o^2}{2g}

The object referred to in the question is thrown from a height H=0 and the maximum height is hm=77.5 m.

(a)

To find the initial speed we solve for vo:

\displaystyle v_o=\sqrt{2gh_m}

v_o=\sqrt{2\cdot 9.8\cdot 77.5}

v_o=39\ m/s

(b)

The maximum time or the time taken by the object to reach its highest  point is calculated as follows:

\displaystyle t_m=\frac{v_o}{g}

\displaystyle t_m=\frac{39}{9.8}

t_m=4\ s

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A rocket, initially at rest on the ground, accelerates straight upward from rest with constant (net) acceleration 29.4 m/s2 m /
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Answer:

The maximum height is 2881.2 m.

Explanation:

Given that,

Acceleration = 29.4 m/s²

Time = 7.00 s

We need to calculate the distance

Using equation of motion

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

Put the value into the formula

s=0+\dfrac{1}{2}\times29.4\times7^2

s=720.3\ m

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Using formula of velocity

v=a\times t

Put the value into the formula

v=29.4\times7

v=205.8\ m/s

We need to calculate the height

Using formula of height

H=\dfrac{v^2}{2g}

Put the value into the formula

H=\dfrac{(205.8)^2}{2\times9.8}

H=2160.9\ m

We need to calculate the maximum height

Using formula for maximum height

H'=H+s

Put the value into the formula

H'=2160.9+720.3

H'=2881.2\ m

Hence, The maximum height is 2881.2 m.

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