The ball is in the air for about 5.5 seconds when it is thrown vertically up.
Acceleration is rate of change of velocity.
[tex]\large {\boxed {a = \frac{v - u}{t} } }[/tex]
[tex]\large {\boxed {d = \frac{v + u}{2}~t } }[/tex]
a = acceleration ( m/s² )
v = final velocity ( m/s )
u = initial velocity ( m/s )
t = time taken ( s )
d = distance ( m )
Let us now tackle the problem!
This problem is about Projectile Motion
Given:
initial speed = u = 27 m/s
Unknown:
time interval of the ball in the air = t = ?
Solution:
[tex]h = u \sin \theta ~t - \frac{1}{2}gt^2[/tex]
[tex]0 = u \sin \theta ~t - \frac{1}{2}gt^2[/tex]
[tex]u \sin \theta ~ t = \frac{1}{2}gt^2[/tex]
[tex]u \sin \theta = \frac{1}{2}gt[/tex]
[tex]t = \boxed {\frac{ 2u \sin \theta }{g}}[/tex]
If the angle of projection = θ = 90° , then :
[tex]t = \boxed {\frac{ 2(27) \sin 90^o }{9.8}}[/tex]
[tex]t \approx 5.5 ~ seconds[/tex]
If the angle of projection = θ = 45° , then :
[tex]t = \boxed {\frac{ 2(27) \sin 45^o }{9.8}}[/tex]
[tex]t \approx 3.9 ~ seconds[/tex]
Grade: High School
Subject: Physics
Chapter: Kinematics
Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle
