Question

Q A projectile's launch speed is 6 times that of its speed at its maximum height. Find the launch angle.

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Answer 1

We adopt the positive direction choices used in the textbook so that equations such as

Equations are directly applicable. The coordinate origin is at its initial position (where it is launched).

At maximum height, we observe v

y

=0 and denote v

x

=v (which is also equalto v

0s

).

In this notation, we have v v 0 = 5 . Next, we observe v

0

cosθ

0

=v

0x

=v,

so that we arrive at an equation (where v



=0 cancels) which can be solved for θ

0

:

$(5v)cosθ </p><p>0</p><p> </p><p> =v⇒θ </p><p>0</p><p> </p><p> =cos </p><p>−1</p><p> ( </p><p>5</p><p>1</p><p> </p><p> )=78.5 </p><p>0</p><p> .$