Question

practice/lab
directions: answer the following questions based on our notes.
part ii - math
a student throws a ball straight upward and it takes 3 seconds to reach the top. answer the following questions.
question | answer

1. calculate the ball’s initial velocity when first thrown. |
2. calculate the ball’s final velocity right before it hits the student’s hand. |
3. calculate the total time it takes the ball to go up and come back down. |

a rock is now dropped from a cliff and takes 30 seconds to hit the ground. calculate the rock’s velocity right before it hits the ground. |
a second rock is dropped from the same cliff and takes 20 seconds to hit the ground. calculate the second rock’s velocity right before it hits the ground. |

Explanation:

Question 6:

Step1: Recall kinematic equation

For vertical motion, at the top, final velocity \( v = 0 \), acceleration \( a=-g=-9.8\ m/s^2 \), time \( t = 3\ s \). Use \( v=u+at \).

Step2: Solve for initial velocity \( u \)

Rearrange \( u=v - at \). Substitute \( v = 0 \), \( a=-9.8\), \( t = 3 \): \( u=0-(-9.8)(3)=29.4\ m/s \).

Step1: Symmetry in vertical motion

Time up = time down (3 s). Use \( v=u+at \) for downward motion (initial velocity at top is 0, \( a = 9.8\ m/s^2 \), \( t = 3\ s \)).

Step2: Calculate final velocity

\( v=0+(9.8)(3)=29.4\ m/s \) (magnitude equal to initial, opposite direction, but if considering speed or magnitude, it's \( 29.4\ m/s \)).

Step1: Time up and down

Time to go up is 3 s, time to come down is also 3 s (symmetry in free - fall when starting and ending at same height).

Step2: Total time

Total time \( T=3 + 3=6\ s \).

Answer:

\( 29.4\ m/s \)

Question 7: