Question #b8b35

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Question #b8b35

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Answer 1 · 2017-02-19T22:58:50

Because the vertical velocity is constantly changing under the influence of gravity and we are not initially sure where the trajectory peaks, the most efficient way to find the initial velocity in this question is to start from the horizontal velocity.

Explanation:

The issue is that the vertical velocity changes (the horizontal velocity stays constant). By the time the ball reaches the wall, the initial vertical velocity has decreased quite a bit. It may also have gone to zero (if the ball is at the peak of its trajectory) or changed direction (if it has started to move downward after passing the peak.) If you try to use this value to find the initial velocity, it will not yield the correct answer.

You've done the question correctly by using the horizontal velocity to find the initial velocity. You can then use that to find the vertical velocity.

What you can't do is simply divide the 7.2 m height of the roof by the 2.1 s and then use that to calculate the initial vertical velocity using the sin.

Let's work out the initial vertical velocity, $v_{y 1}$ $v_(y1)$ from your calculated initial velocity of $2.0 \times 10^{1}$ $2.0 \times 10^1$ at $53^{o}$ $53^o$ to the horizontal:

$v_{y 1} = v sin θ = 20 \times {sin 53}^{o} = 15.97$ $v_(y1) = v sin \theta = 20 \times sin 53^o = 15.97$ $m s^{- 1}$ $ms^-1$

Given the precision to which we were given the data, this should be written as $v_{y 1} = 16$ $v_(y1) = 16$ $m s^{- 1}$ $ms^-1$ or $v_{y 1} = 1.6 \times 10^{1}$ $v_(y1) = 1.6 \times 10^1$ $m s^{- 1}$ $ms^-1$

Now we can use this to calculate the maximum height to which the ball flew. For the top of the trajectory $v_{y} = 0$ $v_y = 0$ :

$v^{2} = u^{2} + 2 a s$ $v^2 = u^2 + 2as$

$v_{y}^{2} = v_{y 1}^{2} + 2 a s$ $v_(y)^2 = v_(y1)^2 +2as$

$0 = 16^{2} + 2 (- 9.8) s$ $0 = 16^2 + 2(-9.8)s$

Rearranging, $s = 13.06$ $s = 13.06$ $m$ $m$

That's significantly higher than the $7.2$ $7.2$ $m$ $m$ of the wall.

Now let's find out when it reaches that height: before or after crossing the wall.

$v = u + a t$ $v = u + at$

$v_{y} = v_{y 1}^{2} + a t$ $v_(y) = v_(y1)^2 + at$

$0 = 16 + (- 9.8) t$ $0 = 16 + (-9.8)t$

Rearranging, $t = 1.63$ $t = 1.63$ $s$ $s$ .

This is less than the $2.1$ $2.1$ $s$ $s$ it takes to travel from the point of the kick to the wall, so the ball rises higher than the wall.

I hope this helps to answer your question: the ball does not simply move from $0$ $0$ $m$ $m$ to $7.2$ $7.2$ $m$ $m$ in a straight line in the $2.1$ $2.1$ $s$ $s$ . Rather, in the vertical direction it rises to $13.06$ $13.06$ $m$ $m$ high then falls back to $7.2$ $7.2$ $m$ $m$ .

Side note: It's better to write $2.0 \times 10^{1}$ $2.0 \times 10^1$ than just $2.0 \times 10$ $2.0 \times 10$ - more correct scientific notation.

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Question #b8b35

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