Question

The maximum speed with which you can throw a stone is about 25 m/s. Can you hit a window 50 m away and 13 m up from the point where the stone leaves your hand? What is the maximum height of a window you can hit at this distance?

Answer 1

Answer:

no, the maximum height = 12.3m

Explanation:

The equation for the trajectory of the ball for a given angle can be derived from:

(1) $x=cos\phi v_0t$

and

(2) $y=-\frac{1}{2}gt^2 +sin\phi v_0t$

Combining equation 1 and 2:

(3) $y=xtan\phi -\frac{gx^2}{2cos^2\phi v_0^2}$

Taking the derivative:

(4) $\frac{dy}{d\phi}=\frac{xsec^2\phi}{v_0^2}(v_0^2-gxtan\phi)$

Setting equation 4 equal to zero to calculate the maximum angle:

(5) $\phi=tan^{-1}(\frac{v_0^2}{gx})$

Plugging equation 5 back into equation2 confirms, the trajectory can't reach the target. The maximum height is 12.3m.

Answer 2

Answer:

;)

Explanation: