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

;)

Step-by-step explanation:

Answer 2

no, the maximum height = 12.3m

Step-by-step 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=-(1)/(2)gt^2 +sin\phi v_0t`

Combining equation 1 and 2:

(3)
`y=xtan\phi -(gx^2)/(2cos^2\phi v_0^2)`

Taking the derivative:

(4)
`(dy)/(d\phi)=(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)((v_0^2)/(gx))`

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