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4 years ago

Calculate the height from which a body is released from rest if its velocity just before hitting the ground is 30ms

1 answer:

6 0

S= ?
U= 0 $ms^{-1}$
V= 30 $ms^{-1}$
A= 9.8 $ms^{-2}$
T=

$V^2 = U^2 + 2AS$
$900 = 19.6S$
$\frac{900}{19.6} = S$
S = 45.92 $m$

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3 years ago

An egg is thrown nearly vertically upward from a point near the cornice of a tall building. The egg just misses the cornice on t

diamong [38]

Answer:

1. 18.5m/s

2. 17.5 m

3. 0 at its highest point

4. Direction is downwards

Explanation:

1. This egg is thrown vertically from a height

Yo = 0. This egg then falls to the point y = -30.0 at t = 5seconds

Y-Yo = V0t - 1/2gt²

-30-0 = V0(5)-1/2(9.8)(5²)

-30 = 5v0 - 4.9x25

-30 = 5V0 - 122.5

-30+122.4 = 5v0

V0 = 92.5/5

= 18.5m/s

this is the initial speed of the egg

2. When the egg is at a maximum height it would have a velocity equal to 0

V² = V0² - 2*g*y

V = 0, V0 = 18.5, g = 9.8

0 = 18.5²-2x9.8*y

342.25-19.6y = 0

342.25 = 19.6y

Divide through by 19.6

Y = 342.25/19.6

Y = 17.5m

this value is how high it rises above starting point

The magnitude of velocity is = 0 at its highest point

This egg falls under gravity. Therefore the acceleration due to gravity has a constant magnitude and direction. Magnitude = 9.8m/s and it's direction is downwards.

5. Please check attachment for graph