1. A clam dropped by a seagull takes 3.0 seconds to hit the ground. What is the seagull’s approximate height above the ground at the time the clam was dropped? 15 m 30. m 45 m 90. m
2.A cannon with a muzzle velocity of 500. meters per second fires a cannonball at an angle of 30.° above the horizontal. What is the vertical component of the cannonball’s velocity as it leaves the cannon? 0.0 m/s 250. m/s 433 m/s 500. m/s
what answers did you get and how do i do it? what formals do i use? ur help is greatly appreciated!
The document has moved here.
For Question 1. We have three equation: v = u + at s = ut + 1/2at^2 (^ is raise to power) v^2 – u^2 = 2as
Here v = final velocity u = initial velocity a = acceleration t = time s = distance
You must remember above three equation.
For above problem acceleration is gravity i.e. g = 9.8m^2/s initial velocity u = 0 time t = 3 sec
We apply second equation s = 0 x 3 + 1/2 x 9.8 x 3 s = 1/2 x 9.8 x 3 s = 14.7 m
For second problem Use trigonometry equation
/ / / 500 m/s / /_30 deg_____________
For horizontal velocity sin 30 deg = v(hor)/v or v(hor) = v x sin 30 deg = 500 x (square root 3)/2 = 433 m/s
For vertical velocity cos 30 deg = v(vert)/v or v(vert) = v x cos 30 deg = 500 x 1/2 = 250 m/s
Name (required)
Mail (will not be published) (required)
Website
