Question

A boy uses a slingshot to launch a pebble straight up into the air. The pebble reaches a height of 37.0 m above the launch point 2.3 seconds later. Assume air resistance is negligible (a) What was the pebble's initial speed (just after leaving the slingshot)? m/s (b) How much time did it take for the pebble to first reach a height of 18.5 m above its launch point? s

Answer 1

Answer:

V0=27.4 m/s; t=0.8 s

Explanation:

Final position y=37.0 m, time = 2.3 s; Initial position is set to be zero. We calculate the initial speed with the kinematics equation:

$y_f=v_0t-0.5*g*t^2$ We solve for initial speed

$v_0=\frac{y_f+0.5gt^2}{t}=\frac{37+4.9*2.3^2}{2.3}=27.4m/s$

Now, using the same expression we estimated time to first reach 18.5 m :

$18.5=27.4t-4.9t^2$ Second order equation with solutions

t1=0.8 s and t2=4.8 s

The first time corresponds to the first reach.