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Brut [27]
3 years ago
11

A football game begins with a coin toss to determine who kicks off. The referee tosses the coin upward with an initial speed of

5 m/s, ignore air resistance.
a)How high does the coin go above the point of release?
b)What is the total time the coin is in the air before returning to the release point?
Physics
1 answer:
erastova [34]3 years ago
5 0

Answer:

(a) 1.28 m

(b) 0.51 s

Explanation:

Given:

Initial speed, U = 5 m/s

Consideration;

1. When the coin is tossed upward, the coin is moving against gravity. Therefor, the acceleration due to gravity experienced by the coin is negative, -g = -9.8 m/s².

2. The final velocity of the coin in the air is zero.  

From equation of motion under free fall,

                           V² = U² + 2gh

                            h = h = \frac{V^{2} - U^{2}}{2(-g)}

Where;

V is the final velocity

U is the initial velocity

g is the acceleration due to gravity

h is the height

                              h = \frac{0^{2} - 5^{2}}{2(-9.8)}

                              h = \frac{0^{-25}{19.60}

                              h = 1.2755 m

                              h = 1.28 m  

(b) The coin still has its initial velocity, u to be equal to 5 m/s, final velocity, v to be equal to 0 m/s and acceleration due to gravity to be -9.8 m/s while in the air before returning to the release point.

From equation of motion under free fall,

                                     V = U + gt

                                     t = \frac{V - U}{g}

                                     t = \frac{0 - 5}{-9.8}

                                     t = 0.51 s

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The distance the package above the ground when it was released, s ≈ 530 meters

<h3 /><h3>What are kinematic equations?</h3>

The kinematic equation of motion gives the interrelationships of the variables of motion.The correct option for the distance the package above the ground when it was released, is the third option;

It is given that:

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The time it takes the package to strike the ground = 12 seconds

The required parameter:

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From the symmetry of the motion of a projectile, the velocity of the package when returns to its staring point where it was dropped = u (Downwards) = 15 m/s

The distance the package falls, s, which is the distance the package above the ground when it was released, is given as follows;

s = u·t + (1/2)·g·t²

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The distance the package falls, s ≈ 530 m = The height of the

The distance the package above the ground when it was released, s ≈ 530 meters

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