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A question asks to calculate "How long the ball was in the air" and provides following data points: $$ initial\ velocity\ \ v_0=10ms^{-1}\\ Angle\ of\ Launch\ = 45^\text{o}\\ Horizontal\ distance \ travelled = 15m$$

The time of flight formula does not utilize range and yields: $$\frac{2*v_y}{g}=\frac{2*10*sin 45^\text{o}}{9.8} = 1.4s$$

While I completely understand the above formulae and reasoning however since the horizontal component has constant velocity I calculated in below manner: $$v_x*t=Range=> 10*cos 45^\text{o}=15=>t=2.12s$$

Is the question incorrect or I am missing something?

Rohan Frederick
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    Question is wrong. The given horizontal distance is more than what would be covered in $1.4$ second. – Math Lover Oct 27 '20 at 12:26
  • The only way to get that much horizontal distance with those initial conditions is if the landing point is much lower than the launch point or we have a much smaller value of $g$. In either case the application of the $2v_y/g$ formula is incorrect (either wrong formula entirely, or wrong value of $g$). – David K Oct 27 '20 at 12:57
  • Thanks David/MathLover, So in a correctly formed question we can find the value of t from both the vertical component and horizontal component i.e. both the methods I stated should yield the same result right? – Rohan Frederick Oct 27 '20 at 13:53

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