41. Following the hint, we have the time-reversed problem with the ball thrown from the ground, towards the right, at 60° measured counterclockwise from a rightward axis. We see in this time-reversed situation that it is convenient to use the familiar coordinate system with +xas rightwardand with positive angles measured counterclockwise. (a) The x-equation (with x0= 0 and x= 25.0 m) leads to 25.0 m = (v0cos 60.0°)(1.50 s), so that v0= 33.3 m/s. And with y0= 0, and y= h> 0 at t= 1.50 s, we have yy vt gty−=−00122where v0y= v0sin 60.0°. This leads to h= 32.3 m.
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