Asked by judah
The path of a baseball relative to the ground can be modelled by the function f(t)=−t^2+8t+1, where d(t) represents the height of the ball in metres, and t represents time in seconds. What is the speed of the ball when it hits the ground?
Answers
Answered by
judah
i found the time at which it hits the ground = 8.123 by doing quadratic formula so now do i plug that into the equation to find the speed?
Answered by
Steve
when does it hit the ground?
-t^2+8t+1 = 0
Now use that in
Vy(t) = -2t+8
to get the vertical speed. The horizontal speed is constant at
Vx(t) = 8cos(arctan 8) ≈ 1 m/s
The final speed is √(Vx^2 + Vy^2)
-t^2+8t+1 = 0
Now use that in
Vy(t) = -2t+8
to get the vertical speed. The horizontal speed is constant at
Vx(t) = 8cos(arctan 8) ≈ 1 m/s
The final speed is √(Vx^2 + Vy^2)
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