documents into his contact's open car which is traveling 155 km/h on a level highway 78.0 m below. At what angle (to the horizontal) should the car be in his sights when the packet is released???
2006-09-19 17:34:43 · 2 answers · asked by bulldawg771 1 in Science & Mathematics ➔ Physics
49.5 degrees.
First convert everything into m/s.
Then note that you can reduce this problem to a rock being thrown horizontally off a 78m cliff at 215-155 = 60 km/h = 16.67m/s.
Find how long the drop takes, then calculate the horizontal distance traveled by the rock.
Once you have the horizontal distance, it's just a matter of using it with the height through simple trigonometry to find the angle.
2006-09-19 17:49:51 · answer #1 · answered by Epicarus 3 · 0⤊ 0⤋
time taken for packet to fall 78m; 78=1/2x g x t^2 so t= SQR(78/5)=3.95 secs . The packet is overtaking the car at 60 kph=16.67m/s, so the car should be 16.67x 3.95=65.8 metres ahead of chopper and 78 m below so tan of angle=78/65.8=1.185 so the car is 50 degrees below the horizontal as seen from the chopper.
2006-09-19 18:04:14 · answer #2 · answered by zee_prime 6 · 2⤊ 1⤋