A man is deer hunting by a cliff that is 1000.0 m tall. He fell asleep while hunting and woke up at night. He decided to shoot off a few rounds to try and get the attention of someone. If he aims 65 degrees above the horizontal and fires a bullet that leaves his gun at 410.00 m/s over the cliff how long will the bullet rise?
How far above ground level will the bullet be at the pinnacle?
How long does it take the bullet to reach the impact point?
What are the x & y components of the bullet's velocity just prior to striking the impact point?
What is the bullet's net velocity (speed and direction) at the impact point?
How far from the base of the cliff does the bullet land?
Note 1. From release to impact the bullet is in free fall. During this time, only gravity exerts a force, downward, on the bullet. Downward acceleration ay= - 9.8 m/s^2, and since there is no horizontal (x direction) force, ax=0.
Note 2. The motion equations are applied separately for vertical motion with acceleration, and for horizontal motion with acceleration 0.
Note 3. We take the ground below the cliff as the origin of an x,y axis system.
Note 4. The initial ...
Solution includes formula, explanation and calculations. It answers multiple questions regarding deer hunting kinematics in 385 words.