A ball thrown into the air lands on the same horizontal level, 42 m away, and 2.9 s later. Find the magnitude of the initial velocity.

Question

loser66 · Answer

you have the formula for $v = \dfrac{d}{t}$ so v = 42/ 2.9 = 14.48 m/s
If your prof asks you to round it or he considers about significant figure, you must let it = 15.0 m/s ( 2 significant figure)

ybarrap · Answer

You can also use SUVAT, the equations of motion. We know if you throw something up at some point it will begin to come down. So it will start with an initial velocity, decelerate because of gravity, come to a stop and then begin to pick up speed again as it falls, reaching it's initial velocity when it hits the ground. The time it takes to reach maximum height is 1/2 the time of travel. So, max height occurs at 2.9/2 seconds. The SUVAT equation we need is $v=u+ gt$, where $g=-9.81 m/s^2$, $u$ is the initial $vertical$ velocity and $v$ is the final velocity. Therefore, $0=u- 9.81(2.9/2)\implies u=14.2245 $ m/s. This is just the vertical velocity. The horizontal velocity was computed by @loser66. The magnitude of the velocity then is $v_{total}=\sqrt{14.2^2+14.5^2}=\bf 20.3$ m/s. If you also wanted to know the angle it was thrown, you can use $\tan\theta=\frac{14.2}{14.5}\implies\theta=\bf 44.4^\circ$. Note that there was no deceleration in the horizontal direction because we ignored the effects due to air drag. Velocity at the beginning is the same as it was when the ball landed. http://en.wikipedia.org/wiki/Equations_of_motion