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A tennis player standing 12.6m from the net hits the ball at 3.00 degrees above the horizontal. To clear the net, the ball must rise at least 0.330m. If the ball just clears the net at the apex of its trajectory, how fast was the ball moving when it left the racket? Please show wooooorkkkkk =)

User Gjsalot
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2 Answers

3 votes

Final answer:

To determine the initial speed of a tennis ball when it just clears the net, one needs to analyze projectile motion, considering both horizontal and vertical components of the motion, applying kinematic equations and trigonometric functions.

Step-by-step explanation:

A student has asked how fast a tennis ball was moving when it left the racket if it clears the net at 12.6 meters away with a minimum rise of 0.330 meters and a launch angle of 3.00 degrees above the horizontal. To solve this, we need to analyze the projectile motion of the ball and use kinematic equations.

Firstly, we use the equation for the vertical motion of the projectile: y = v_{0y}t + (1/2)gt^2 where y is the vertical displacement, v_{0y} is the initial vertical velocity, t is the time, and g is the acceleration due to gravity (9.81 m/s^2). The ball needs to climb 0.330 m above its starting point to clear the net.

To find t, we use the horizontal motion equation: x = v_{0x}t where x is the horizontal displacement and v_{0x} is the initial horizontal velocity. Since x equals 12.6 m and v_{0x} can be found using the initial velocity v_0 and the launch angle with v_{0x} = v_0 imes cos( heta), we can solve for t.

After isolating the time t from the horizontal equation and substituting it into the vertical equation, we can then solve for the initial speed v_0. This calculation involves trigonometric functions and the quadratic formula. Finally, we'll use the law of sines or cosines, as needed, to find the initial velocity of the ball.

User Oatsoda
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7.0k points
5 votes

Answer:

48.6 m/s

Step-by-step explanation:

height, h = 0.33 m

angle of projection, θ = 3°

Let the velocity of projection is u.

Use the formula for the maximum height raised


H=(u^(2)Sin^(2)\theta )/(2g)


0.33=(u^(2)Sin^(2)3 )/(2* 9.8)

u = 48.6 m/s

User Zambari
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