Question

Sam tosses a ball horizontally off a footbridge at 4.3 m/s. How much time passes after he releases it until its speed doubles?

Can someone please explain the correct equation to use thoroughly and help me to find the correct answer?!

Answer 1

by speed, do you mean the vector sum of the horizontal and vertical speeds?

Must be, since the horizontal speed never changes, and the initial vertical speed was 0, so double that is still 0.

<b>v</b> = 4.3<b>i</b> - 9.8t<b>j</b>
|<b>v</b>| = √(4.3^2 + (9.8t)^2) = √(96.04t^2 + 18.49)

so, double the original speed is 8.6m/s and we want to find t when

8.6 = √(96.04t^2 + 18.49)
t = 0.76 sec

Answer 2

This is the projectile thrown with horizontal velocity.
So
v(x) = const,
v(y) = g•t.
v ² = v(x)² +v(y)²
v =2•v(x),
[2v(v)]² = v(x)² +v(y)²,
3•v(x)² =v(y)² = (g•t)²,
t =v(x) •√3/g =4.3•1.73/9.8 =0.76 s.