Answer:
Option E is correct.
Time the ball remains in the air before striking the ground is closest to 3.64 s
Explanation:
yբ = yᵢ + uᵧt + gt²/2
yբ = 0
yᵢ = 2 m
uᵧ = u sinθ = 20 sin 60 = 17.32 m/s
g = -9.8 m/s², t = ?
0 = 2 + 17.32t - 4.9t²
4.9t² - 17.32t - 2 = 0
Solving the quadratic equation,
t = 3.647 s or t = -0.1112 s
time is a positive variable, hence, t = 3.647 s. Option E.
The time the ball remains in the air before striking the ground is 3.53 seconds.
Projectile motion is the motion experienced by an object thrown into space and it moves along a curved path.
The time (T) the ball remains in the air before striking the ground is given by:
[tex]T=\frac{2usin(\theta)}{g}[/tex]
where u is the velocity = 20 m/s, g is the acceleration due to gravity = 9.81 m/s², θ = angle = 60 degree
[tex]T=\frac{2usin(\theta)}{g}=T=\frac{2*20sin(60)}{9.81}=3.53 \ s[/tex]
Hence the time the ball remains in the air before striking the ground is 3.53 seconds.
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