Question

During a baseball game, a batter hits a high pop-up. if the ball remains in the air for a total of 6.0 s, how high does it rise? ( Assume Accel. = 10 m/s^2 )

Answer 1

• Time=6s
• Vertex height's time= 6/2=3s
• Acceleration due to gravity=10m/s^2

Apply second equation of kinematics

$\\ \rm\hookrightarrow s=ut+\dfrac{1}{2}gt^2$

$\\ \rm\hookrightarrow s=1/2(10)(3)^2$

$\\ \rm\hookrightarrow s=5(9)$

$\\ \rm\hookrightarrow s=45m$

It rises 45m