# Answer to Question #7018 in Mechanics | Relativity for tanmay

Question #7018

The benches of gallery in cricket ground are 1m wide and 1 m high.A batsman strikes the ball at a level 1 m above the ground and hits a sixer.The ball starts at 35m/s at an angle of 53degree with horizontal.The benches are perpendicular to the plane of motion and the first bench is 110m from batsman.ON WHICH BENCH WILL THE BALL HIT?

Expert's answer

One way of doing this is to find an equation for the trajectory of the ball. We'll also find an equation for the benches.

We'll interpret the width of the benches as being the distance from the front to back. For the purpose of this question we can take the benches to be a straight line that goes up 1 m for every 1 m back. You know how a stadium is right, the seats further back are higher. It's not true that the seats at the back are at the same height at the seats in front, otherwise the fans in the back will see nothing!

So, we know is line will have a gradient of 1. If we take our origin to be the point where the ball was hit.then the benches will follow the line y = x - 110.

Now we use the kinematic equations of motion to find an equation for the (parabolic) trajectory the ball traces out. We know it will be an equation of the form y = a(x-h)² + k, where k is the height of the highest point and h is the horizontal coordinate of that point. We also know, from the kinematic equation in the vertical direction d = ut + 1/2 gt², that a = g/2 = -9.8/2 = -4.9.

So the parabola is y = -4.9(x-h)² + k.

Let's find the values of k and h now.

k first: We use the kinematic equations of motion. We know this is the height of the highest point, and at this point, the vertical component of the speed is 0 m/s since it will have stopped going up but not yet begun coming down.

To find k:

u = initial velocity = 35sin53 = 27.95 m/s

v = final velocity = 0 m/s

k = vertical distance = ?

a = acceleration due to gravity = -9.8 m/s²

v² = u² + 2ak, so since v² = 0, k = -u²/2a, and plugging in the numbers gives k = 39.86 m.

To find t, the time parameter which will help find h:

v = u + at, so t = -u/a = 2.85 s.

Now we turn our attention to the horizontal direction to find h. This will be the horizontal coordinate at time 2.85 s. We know that the only force acting on the ball is the purely vertical force of gravity, so there are no horizontal forces acting on the ball, therefore no horizontal acceleration and therefore the horizontal component of velocity is constant. And the horizontal component of velocity is 35cos53. So horizontal distance = horizontal velocity times time = (2.85 s)(35cos53 m/s) = 60.07 m.

So the trajectory of the ball is given by

y = -4.9(x - 60.07)² + 39.86, which can be written

y = -4.9(x² - 120.14x + 3609) + 39.86, or

y = -4.9x² + 589x -17644

The point where the ball hits the benches will be where this line intersects with the line y = x - 110.

So we solve -4.9x² + 589x - 17644 = -x - 110, or

-4.9x² + 588 x + 17754 = 0, a quadratic equation with solutions obtained using the quadratic formula of 145 and a negative solution we may disregard.

So the y-coord is y = x - 110 = 145 - 110 = 35 m high, so it hits the 35th bench.

We'll interpret the width of the benches as being the distance from the front to back. For the purpose of this question we can take the benches to be a straight line that goes up 1 m for every 1 m back. You know how a stadium is right, the seats further back are higher. It's not true that the seats at the back are at the same height at the seats in front, otherwise the fans in the back will see nothing!

So, we know is line will have a gradient of 1. If we take our origin to be the point where the ball was hit.then the benches will follow the line y = x - 110.

Now we use the kinematic equations of motion to find an equation for the (parabolic) trajectory the ball traces out. We know it will be an equation of the form y = a(x-h)² + k, where k is the height of the highest point and h is the horizontal coordinate of that point. We also know, from the kinematic equation in the vertical direction d = ut + 1/2 gt², that a = g/2 = -9.8/2 = -4.9.

So the parabola is y = -4.9(x-h)² + k.

Let's find the values of k and h now.

k first: We use the kinematic equations of motion. We know this is the height of the highest point, and at this point, the vertical component of the speed is 0 m/s since it will have stopped going up but not yet begun coming down.

To find k:

u = initial velocity = 35sin53 = 27.95 m/s

v = final velocity = 0 m/s

k = vertical distance = ?

a = acceleration due to gravity = -9.8 m/s²

v² = u² + 2ak, so since v² = 0, k = -u²/2a, and plugging in the numbers gives k = 39.86 m.

To find t, the time parameter which will help find h:

v = u + at, so t = -u/a = 2.85 s.

Now we turn our attention to the horizontal direction to find h. This will be the horizontal coordinate at time 2.85 s. We know that the only force acting on the ball is the purely vertical force of gravity, so there are no horizontal forces acting on the ball, therefore no horizontal acceleration and therefore the horizontal component of velocity is constant. And the horizontal component of velocity is 35cos53. So horizontal distance = horizontal velocity times time = (2.85 s)(35cos53 m/s) = 60.07 m.

So the trajectory of the ball is given by

y = -4.9(x - 60.07)² + 39.86, which can be written

y = -4.9(x² - 120.14x + 3609) + 39.86, or

y = -4.9x² + 589x -17644

The point where the ball hits the benches will be where this line intersects with the line y = x - 110.

So we solve -4.9x² + 589x - 17644 = -x - 110, or

-4.9x² + 588 x + 17754 = 0, a quadratic equation with solutions obtained using the quadratic formula of 145 and a negative solution we may disregard.

So the y-coord is y = x - 110 = 145 - 110 = 35 m high, so it hits the 35th bench.

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