Question

You are traveling in a convertible with the top down. The car is moving at a constant velocity 22.5 m/s, due east along flat ground. You throw a tomato straight upward at a speed of 10.6 m/s. How far has the car moved when you get a chance to catch the tomato?

Answer 1

Answer:

The car moved a distance $x=48.6m$ .

Explanation:

First we need to know: How much time will the tomato spend in the air?

From Kinematics: $v_y(t)=v_0_y+at$

where $v_0_y=10.6\frac{m}{s}$ and $a=-g=-9.8\frac{m}{s^2}$ is gravity's acceleration.

$v_y(t)=v_0_y-gt$

When the tomato touches the car again, $v_y=-v_0_y=-10.6\frac{m}{s}$

Then, we have:

$t=\frac{v_y-v_o_y}{-g}$ ⇒ $t=\frac{v_y-v_o_y}{-g}=2.16s$

Also from Kinematics we have: $x(t)=v_xt$

Which is very simple because we can take initial position 0 and there's no acceleration in the x direction. And $v_x=22.5\frac{m}{s}$

So, taking $t=2.16s$

$x=48.6m$