Question

a ball is hit with 30 m/s velocity creating an angle of 20 degree.at the place , where the ball reaches it's maximum height, at that point a fielder who is 5 feet and 8 inches tall is fielding.he jumps 2 m high to catch the ball.the distance of the boundary is 70 m from the hitting place.and another fielder is fielding inside 11 m from boundary .

a) will the 1st fielder be able to catch the ball??

b)if the 1st fielder fails to take the catch by any chance , will the 2nd fielder be able to take the catch?

c) does the height of the 1st fielder is necessary to solve part a) ??

Answer 1

Answer on Question #50526, Physics, Mechanics | Kinematics | Dynamics

A ball is hit with $30\mathrm{m / s}$ velocity creating an angle of 20 degree. At the place, where the ball reaches its maximum height, at that point a fielder who is 5 feet and 8 inches tall is fielding. He jumps $2\mathrm{m}$ high to catch the ball. The distance of the boundary is $70\mathrm{m}$ from the hitting place and another fielder is fielding inside $11\mathrm{m}$ from boundary.

a) Will the 1st fielder be able to catch the ball??

b) If the 1st fielder fails to take the catch by any chance, will the 2nd fielder be able to take the catch?

c) Does the height of the 1st fielder is necessary to solve part a) ??

Solution:

Before we start to solve the given problem we will represent the graph of the trajectory of the moving ball.

First, determine the horizontal and vertical components of the initial velocity. We begin by looking at the motion in the y direction. We have the following.

$$v _ {0 y} = v _ {0} \sin \theta = \left(3 0 \frac {m}{s}\right) \cdot (\sin 2 0 {}^ {\circ}) = 3 0 \cdot 0. 3 4 2 = 1 0. 2 6 1 m / s$$ $$v _ {0 x} = v _ {0} \cos \theta = \left(3 0 \frac {m}{s}\right) \cdot (\cos 2 0 {}^ {\circ}) = 3 0 \cdot 0. 9 3 9 7 = 2 8. 1 9 1 m / s$$ $$a = g = - 9. 8 1 \frac {m}{s ^ {2}}$$

Now we need to determine the time it took to land.

$$y = v _ {0 y} t - \frac {1}{2} g t ^ {2}$$

Since $y = 0$ at the end of flight we have:

$$0 = \left(10.261 \frac{\mathrm{m}}{\mathrm{s}}\right) \mathrm{t} - \frac{1}{2} \left(9.81 \frac{\mathrm{m}}{\mathrm{s}^{2}}\right) \mathrm{t}^{2}$$

We solve the obtained equation for $t$.

$$\left(\left(10.261 \frac{\mathrm{m}}{\mathrm{s}}\right) - \frac{1}{2} \left(9.81 \frac{\mathrm{m}}{\mathrm{s}^{2}}\right) \mathrm{t}\right) \mathrm{t} = 0$$

Simplify the equation.

$$\left(10.261 \frac{\mathrm{m}}{\mathrm{s}}\right) - \frac{1}{2} \left(9.81 \frac{\mathrm{m}}{\mathrm{s}^{2}}\right) \mathrm{t} = 0$$ $$\left(10.261 \frac{\mathrm{m}}{\mathrm{s}}\right) - 4.905 \frac{\mathrm{m}}{\mathrm{s}^{2}} \mathrm{t} = 0$$

Then we can find the time of the ball in air.

$$4.905 \frac{\mathrm{m}}{\mathrm{s}^{2}} \mathrm{t} = 10.261 \frac{\mathrm{m}}{\mathrm{s}}$$

Divide both sides by 4.905.

$$t = \frac{10.261 \frac{\mathrm{m}}{\mathrm{s}}}{4.905 \frac{\mathrm{m}}{\mathrm{s}^{2}}}$$ $$t = 2.092 \mathrm{s}$$

Now we can find the distance the ball travels before it hits the ground if we assume that the first fielder will not try to catch the ball.

$$x = v_{x} t$$

Substitute into the formula the find values.

$$x = 28.191 \mathrm{m/s} \cdot 2.092 \mathrm{s}$$ $$x = 58.976 \mathrm{m}$$

This means the ball will travel $58.976\mathrm{m}$.

Then we have to find maximum height. Firstly calculate the half the flight time, which will be equal to 1.046 seconds. Substitute the find value into the following formula.

$$y = (10.261 \frac{\mathrm{m}}{\mathrm{s}}) (1.046 \mathrm{s}) - \frac{1}{2} (9.81 \frac{\mathrm{m}}{\mathrm{s}^{2}}) (1.046 \mathrm{s})^{2}$$

We simplify the equation by opening the parenthesis.

$$y = (10.261 \frac{\mathrm{m}}{\mathrm{s}}) (1.046 \mathrm{s}) - \frac{1}{2} (9.81 \frac{\mathrm{m}}{\mathrm{s}^{2}}) (1.046 \mathrm{s})^{2}$$ $$y = 10.733 \mathrm{m} - 5.367 \mathrm{m}$$ $$y = 5.366m$$

The ball will reach a maximum height of 5.366 meters.

Thus we can conclude that, on the basis of our calculations, we found the coordinates at which the first player to be able to catch the ball, it must be located at a distance of about 25 meters from the place where the ball was thrown, if we assume that he is at the point where the ball reaches maximum altitude, in this case the player is not able to catch him, because the maximum height is 5.37 meters, and the player jumps up to a height of 2 meters, if we consider the second player, we can conclude that he can catch the ball because the ball travels a distance approximately equal to 59 meters, according to the statement of the problem, the second player is placed at a distance of 11 meters from the border that is 70 meters, which is 59 meters.

The height of the player in the first problem does not affect the settlement of accounts because the importance is the maximum height that reaches the player in jump to catch the ball that moves along a predetermined path.

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