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(a)An archer shoots an arrow from a height of 1.21 m above ground with an initia

ID: 1956074 • Letter: #

Question

(a)An archer shoots an arrow from a height of 1.21 m above ground with an initial velocity of 47.5 m/s and an initial angle of 36.5° above the horizontal. At what time after the release of the arrow from the bow will the arrow be flying exactly horizontally?

(b)You serve a tennis ball from a height of 2.11 m above the ground. The ball leaves your racket with a speed of 19.7 m/s at an angle of 6.57° above the horizontal. The horizontal distance from the court's baseline to the net is 11.89 m, and the net is 1.07 m high. Neglect spin imparted on the ball as well as air resistance effects. Does the ball clear the net (= positive answer)? If yes, by how much? If not, by how much did it miss? In that case the answer will be negative.

Explanation / Answer

(a) The vertical position of the arrow can be described by the equation:

y = 1.21 + 47.5 Sin(36.5o) t - (1/2)g t2

Where 1.21 is the initial hight, 47.5 Sin(36.5o) is the y-component of the initial velocity, and g is gravity.

When the arrow is flying exactly horizontally, the vertical velocity of the arrow will be 0. We can take the derivative of the above equation to find the y-velocity as a function of time:

vy = 47.5 Sin(36.5o) - gt

Set it equal to 0:

0 = 47.5 Sin(36.5o) - gt

t = 2.88s (with g = 9.8) (if g = 10, then t = 2.83s)

(b) The y-position of the ball can be respresented as:

y = 2.11 + 19.7 Sin(6.57o) t - (1/2)g t2

And the x-position:

x = 19.7 Cos(6.57o) t

(there is no acceleration in the x-direction and the initial position is 0)

To get to the net, the ball has to go 11.89 m, so the time it gets there is given by the x equation:

11.89 = 19.7 Cos(6.57o) t

t = 0.608s

Now we want to find the height of the ball at that time, so we plug 0.608 in for t in the y equation:

y = 2.11 + 19.7 Sin(6.57o) (0.608) - (1/2)g (0.608)2

For g = 9.8: y = 1.67m

For g = 10: y = 1.63m

Either way, it did clear the net (both numbers are greater than 1.07, the height of the net)

If you are using g = 9.8, it cleared by 0.6m

And for g = 10, it cleared by 0.56m