Answer:
a) t = 2.55s
b) [tex]v_{0x} = 9.80 m/s[/tex]
c) yes
Explanation:
In order to solve this problem, we can start by drawing a sketch of the situation so we can better visualize what the problem is about (see attached picture).
a)
For the first question. We are talking about a movement in two dimensions. So on the first question they are asking us for vertical movement. It will be uniformly accelerated, so we can use the following formula:
[tex]y_{f}=y_{0}+V_{0y}t+\frac{1}{2}at^{2}[/tex]
We know the following:
[tex]y_{f}=0[/tex]
[tex]y_{0} = 32m[/tex]
[tex]V_{0y}=0[/tex]
t=?
[tex]a=-9.81 m/s ^{2}[/tex]
With this data, we can simplify our equation, so we end up with:
[tex]y_{0}+\frac{1}{2}at^{2}=0[/tex]
so we can now substitute the data we know and solve for t:
[tex]32m-\frac{1}{2}(9.81 m/s^{2})t^{2}=0[/tex]
[tex](-4.905 m/s^{2})t^{2}=-32m[/tex]
[tex]t^{2} = \frac{-32m}{-4.905 m/s^{2}}[/tex]
[tex]t=\sqrt{6.52s^{2}}[/tex]
t = 2.55 s
b)
For part b, since we are talking about horizontal movement and we are neglecting drag, this means that his horizontal speed will be constant. So we can use the following formula:
[tex]V_{x}=\frac{x}{t}[/tex]
we know he most move a horizontal distance of 25 meters in a time of 2.55s so we get:
[tex]V_{x} = \frac{25m}{2.55s}[/tex]
[tex]V_{x}=9.80 m/s[/tex]
c) for part c, we can do the conversion between miles per hour to meters per second like this:
[tex]\frac{27mi}{1hr}*\frac{1hr}{3600s}*\frac{5280ft}{1mi}*\frac{0.3048m}{1ft}[/tex]
so the given initial speed is equivalent to:
12.07 m/s
this is greater than the minimum 9.80 m/s we need, so the skater will clear the pool at this speed.
(a) The time spent in air by the skateboarder is 2.56 s
(b) The initial speed needed by the skateboarder to clear the pool is 9.77 m/s.
(c) The skateboarder with given initial speed of 27 mph will clear the pool since it is greater than minimum initial velocity required.
The given parameters;
The time spent in air by the skateboarder is calculated as follows;
[tex]h = v_0t + \frac{1}{2} gt^2\\\\h = 0 + \frac{1}{2} gt^2\\\\gt^2 = 2h\\\\t = \sqrt{\frac{2h}{g} } \\\\t = \sqrt{\frac{2\times 32}{9.8} } \\\\t =2.56 \ s[/tex]
The initial speed needed by the skateboarder to clear the pool is calculated as;
[tex]v = \frac{d}{t} \\\\v = \frac{25}{2.56} \\\\v = 9.77 \ m/s[/tex]
The initial speed of the skateboarder is m/s;
[tex]27 \ \frac{mile}{hour} \times \frac{1609.3 \ m}{1 \ mile} \times \frac{1 \ hr}{3600 \ s} = 12.07 \ m/s[/tex]
Thus, the skateboarder with given initial speed of 27 mph will clear the pool since it is greater than minimum initial velocity required.
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