KE= 0.5(m)(v^2)
KE= 250,000 J
Answer:
A.) 1.3 seconds
B.) 0.42 m
Explanation:
A.) You are given the angle of projection to be 40 degrees and initial velocity of 20m/s.
At vertical component
U = Usin 40 that is,
U = 20sin40
Using the first equation of motion under gravity
V = U - gt
Let V = 0
0 = UsinØ - gt
gt = UsinØ
t = UsinØ/g
Where U = 20 m/s
Ø = 40 degree
g = 9.8 m/s^2
Substitutes all the parameters into the formula
t = 20sin40/9.8
t = 1.3 seconds
Total time of flight T = 2t
T = 2 × 1.3 = 2.6 s
B.) To calculate the maximum height,
You will use the formula
V^2 = U^2 - 2gH
At maximum height, V = 0
2gH = Usin^2Ø
H = Usin^2Ø/ 2g
Substitutes all the parameters into the formula
H = 20 sin^2(40) ÷ 2(9.8)
H = 8.2635/19.6
H = 0.42 m
The answer is
<em>All the three light bulbs will glow</em>
hope this helps
Answer:
48.51ms / 174.6 km/h
Explanation:
y = 1/2 x g x t^2 v = g x t
when y = 120m
120 = 1/2 x 9.8 x t^2
t^2 = 24.49
t = 4.95s
when t = 4.95s
v = 9.8 x 4.95
v = 48.51 m/s = 174.6 km/h
I'd say its realistic. But I don't really know that sry