Answer to Question #134229 in Classical Mechanics for leydiana n marroquin

Question

During a baseball game, a batter hits a high
pop-up.
If the ball remains in the air for 4.72 s, how
high above the point where it hits the bat
does it rise? Assume when it hits the ground
it hits at exactly the level of the bat. The
acceleration of gravity is 9.8 m/s
2
.
Answer in units of m.

Answer 1

solution

give data

time taken durig whole motion(T)=4.72 sec

during its motion in y-axis it will be at maximum height which can be given as

$h_{max} = u_{y}t -\frac{gt^2}{2}$ ......eq.1

it will take time to attain a maximum height is $\frac{T}{2}$

$t = 4.72/2 =2.36 \; s$

by applying the first equation of motion

$v_y=u_y+gt$

at the maximum height final velocity will be zero

so

$0 = u_{y} - 9.8 \times 2.36$

then

$u_{y} = 23.12m/s$

by puttig the value in equation 1

$h_{max} = 23.12 \times 2.36 - \frac{9.8 \times 2.36^2}{2}$ m

$h_{max}=27.29m$

so height attained by the ball is 27.29 meter .