Answer:
α=0.625rad/s^2
v=340m/s
w=10rad/s
θ=320rad
Explanation:
Constant angular acceleration = ∆w/∆t
angular acceleration = 20/32
α=0.625rad/s^2
Linear velocity v=wr
v = 20×17= 340m/s
Average angular velocity
w0+w1/2
w= 0+20/2
w= 20/2
w=10rad/s
What angle did it rotate with
θ=wt
θ= 10×32
=320rad
Inconsistent. You should take three readings at least.
Answer:
Explanation:
The high reached by a proyectile in an uniformly accelerated motion is given by:
The time that the rocket spends in the air is obtained for y = 0, since this is the time that the rocket travels before touching the ground. Recall that 
. Solving for t:
Answer:
sure if I can figure out how to follow you
Explanation:
recenty joined
Answer:
vx = 65 yd/3 sec = 21.7 yd/sec since horizontal speed is constant
vy = g t = (32 / 3) yd/sec^2 * 1.5 sec = 16 yd/sec where 32/3 is the acceleration due to gravity in yds / sec^2 and 1.5 is the time to travel each way in the vertical direction
V = (vx^2 + vy^2)^1/2 = (21.7^2 + 16^2)^1/2 = 27 yd/sec
tan theta = vy/vx = 16 / 21.7 = .737 theta = 36.4 deg
You can check using the range formula:
R = v^2 sin (2 theta) / g = 27^2 * .955 / (32 / 3) = 65.3 yds
The difference from 65 yds may be rounding error.
