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A ball is thrown from a point 6 metres above the ground. The graph shows the height of the ball above the ground in metres. Estimate the speed of the ball. in m/s after 1 second You must show your working Time (Seconds) 4 5 square 3

Question

A ball is thrown from a point
6 metres above the ground.
The graph shows the height of
the ball above the ground
in metres.
Estimate the speed of the ball.
in m/s after 1 second
You must show your working
Time (Seconds) 4	5
square 
3

A ball is thrown from a point 6 metres above the ground. The graph shows the height of the ball above the ground in metres. Estimate the speed of the ball. in m/s after 1 second You must show your working Time (Seconds) 4 5 square 3

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DaphneVeteran · Tutor for 11 years

Answer

The estimated speed of the ball after 1 second is approximately 0 m/s.

Explain

## Step 1: <br />The problem does not provide a graph, but it does tell us that the ball is thrown from a height of 6 meters. We can assume that the ball is thrown vertically upward. <br /><br />## Step 2: <br />The speed of the ball after 1 second can be estimated using the formula for the velocity of an object under the influence of gravity. This formula is given by:<br /><br />### \( v = u - g \times t \)<br /><br />where:<br />- \( v \) is the final velocity (which we are trying to find),<br />- \( u \) is the initial velocity (which we don't know),<br />- \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)), and<br />- \( t \) is the time (1 second in this case).<br /><br />## Step 3: <br />Since we don't know the initial velocity of the ball, we need to make an assumption to estimate it. A reasonable assumption would be that the ball was thrown with enough velocity to reach its maximum height in about 1 second. This would mean that the initial velocity was approximately equal to the acceleration due to gravity, or \( 9.81 \, \text{m/s} \).<br /><br />## Step 4: <br />Substituting the values into the formula, we get:<br /><br />### \( v = 9.81 \, \text{m/s} - 9.81 \, \text{m/s}^2 \times 1 \, \text{s} = 0 \, \text{m/s} \)
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