A robot probe drops a camera off the rim of a 239 m high , where there acceleration is 9.80 m/s a) Find the velocity with which the camera hits the ground 249.8) (2:4) 46844 Sud 66 mis b) Find the time period for it to hit the ground. 2. A tennis ball is thrown vertically upward with an initial velocity of -8.00 a) What will the ball’s speed be when it returns to its starting point? b) How long will the ball take to research its starting point? 8 14.8)(+)
The Correct Answer and Explanation is :
Let’s analyze the two physics problems you’ve presented, applying the principles of kinematics to find the required solutions.
Problem 1: Free Fall from a Cliff
Given:
- Height of the cliff, ( h = 239 ) meters
- Acceleration due to gravity, ( g = 9.80 ) m/s²
- Initial velocity, ( u = 0 ) m/s (since the camera is dropped)
(a) Velocity upon impact
To determine the velocity with which the camera hits the ground, we use the kinematic equation:
[ v^2 = u^2 + 2gh ]
Plugging in the values:
[ v^2 = 0 + 2 \times 9.80 \, \text{m/s}^2 \times 239 \, \text{m} ]
[ v^2 = 4684.4 \, \text{m}^2/\text{s}^2 ]
Taking the square root of both sides:
[ v = \sqrt{4684.4} \, \text{m/s} ]
[ v \approx 68.44 \, \text{m/s} ]
Therefore, the camera hits the ground with a velocity of approximately 68.44 m/s.
(b) Time to reach the ground
To find the time taken for the camera to hit the ground, we use the equation:
[ v = u + gt ]
Solving for ( t ):
[ t = \frac{v – u}{g} ]
Substituting the known values:
[ t = \frac{68.44 \, \text{m/s} – 0}{9.80 \, \text{m/s}^2} ]
[ t \approx 6.98 \, \text{seconds} ]
Thus, it takes approximately 6.98 seconds for the camera to reach the ground.
Problem 2: Tennis Ball Thrown Vertically Upward
Given:
- Initial velocity, ( u = 8.00 ) m/s
- Acceleration due to gravity, ( g = 9.80 ) m/s² (acting downward)
(a) Speed upon returning to the starting point
When the ball returns to its starting point, its speed will be equal to the initial speed but in the opposite direction (downward). Therefore, the speed is 8.00 m/s.
(b) Time to return to the starting point
The time to reach the maximum height is given by:
[ t_{\text{up}} = \frac{u}{g} ]
[ t_{\text{up}} = \frac{8.00 \, \text{m/s}}{9.80 \, \text{m/s}^2} ]
[ t_{\text{up}} \approx 0.816 \, \text{seconds} ]
Since the time to ascend and descend are equal, the total time to return to the starting point is:
[ t_{\text{total}} = 2 \times t_{\text{up}} ]
[ t_{\text{total}} = 2 \times 0.816 \, \text{seconds} ]
[ t_{\text{total}} \approx 1.63 \, \text{seconds} ]
Therefore, the ball takes approximately 1.63 seconds to return to its starting point.
Explanation:
In both problems, we apply the fundamental equations of kinematics under constant acceleration. For the free-falling camera, the absence of initial velocity simplifies the calculations, allowing us to directly relate the distance fallen to the final velocity and time. In the case of the tennis ball, the symmetry of projectile motion under uniform gravity ensures that the time to ascend to the peak equals the time to descend back to the starting point, and the speed at the starting point upon return matches the initial speed, albeit in the opposite direction. These principles are foundational in physics, illustrating the predictability of motion under constant acceleration.