You are traveling in a convertible with the top down. The car is moving at a constant velocity 20.0m/s due east along flat ground. You throw a tomato straight upward at a speed of 15.9m/s. how far has the car moved when you get a chance to catch a tomato?
Calculate how long the tomato is in the air. Call it T. It's twice the time it takes for the vertical velocity to reach zero, after being thrown. It spends an e
T = 2*(15.9 m/s)/g = 3.24 s
Then multiply that time by 20 m/s.
To solve this problem, we need to consider the horizontal and vertical motion of the tomato separately.
First, let's find out how long it takes for the tomato to reach its maximum height. We can use the kinematic equation for vertical motion:
vf = vi + at
Since the tomato is thrown straight upward, the final vertical velocity (vf) at the highest point would be zero. The initial vertical velocity (vi) is 15.9 m/s, and the acceleration (a) due to gravity is -9.8 m/s^2 (negative because it acts in the opposite direction of motion).
0 = 15.9 - 9.8t
Solving for t, we get:
t = 15.9 / 9.8 ≈ 1.63 seconds
Now, let's determine the horizontal distance the car has traveled during this time. We can use the formula:
d = v * t
where d is the distance, v is the velocity, and t is the time. In this case, the velocity of the car is 20.0 m/s (due east), and the time is 1.63 seconds.
d = 20.0 * 1.63 ≈ 32.6 meters
Therefore, the car has moved approximately 32.6 meters when you get a chance to catch the tomato.