Final answer:
The two spherical balls thrown upwards with the same initial velocity 3 seconds apart will collide at a height of approximately 18.3 meters above the launch point of the second ball.
Step-by-step explanation:
Calculating the Collision Height of Thrown Spherical BallsTo determine the height at which two spherical balls collide when thrown upwards with the same initial velocity, we will use the equations for projectile motion.
Given that the balls are thrown at an interval of 3 seconds and have the same initial velocity of 35 m/s, the first step is to calculate the time it takes for the first ball to reach its peak height.
We use the formula v = u + at, where v is the final velocity (0 m/s at the peak), u is the initial velocity (35 m/s), and t is the time it takes to reach the peak. The acceleration a is due to gravity and will be -9.8 m/s² (negative as it opposes the motion).
Solving for t, we get:
t = 35 / 9.8
t ≈ 3.57 seconds (time to reach the peak)
The second ball is thrown 3 seconds after the first, so there is a difference of around 0.57 seconds in their flight before the first one starts to descend. To find the height at which they collide, we calculate the distance each ball travels in 0.57 seconds. Since the second ball is still ascending, we use:
s = 35 * 0.57 + 0.5 * (-9.8) * (0.57)²
s ≈ 18.3 meters (approximate)
The height of the collision would be the initial height reached by the first ball minus the distance it falls during those 0.57 seconds. By the symmetry of projectile motion, the first ball would fall the same distance the second ball rises in that duration; thus they collide at approximately 18.3 meters above the point of the second throw.