Calculating Initial Velocity & Angle of a Moving Ball

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To calculate the initial velocity and angle of a moving ball, the horizontal and vertical positions are defined by the functions x(t) = 84 + 27t and y(t) = 88 + 46t - 490t². The initial velocity components can be found by differentiating these functions with respect to time, yielding vx = 27 cm/s and vy = 46 cm/s at t = 0. The magnitude of the initial velocity is calculated using the formula vo = sqrt(vx² + vy²). The angle between the initial velocity and the horizontal can be determined using the arctangent of the ratio vy/vx. This approach provides a clear method for solving the problem of the ball's motion.
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A ball is moving through the air as shown. Its horizontal position x and vertical position y can be described by two functions: x(t) = a + b*t and y(t) = c + d*t + e*t*t.
Here t represents time and

a= 84 cm, b=27 cm/s, c=88 cm, d=46 cm/s and e=-490 cm/s/s.

What is the magnitude of the initial velocity of the ball? and what is the angle between the initial velocity and the horizontal?

Could anyone provide a much needed hint to get going on this problem?
 
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Find dx/dt and dy/dt. That will give you vx and vy. Put t = 0 to get vo = sqrt(vx^2 + vy^2)
 
ok i got it thanks so much!
 
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