During a baseball game, a batter hits a high pop-up.
If the ball remains in the air for 5.93 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.

43.01 m

Explanation:

Given that :

Time ball uses in the air = 5.93, time it takes to reach maximum height and return = time of flight(T) = 5.93 s

TIME of flight (T) = 2 * time taken(t)

Where g = a = acceleration due to gravity = 9.8m/s²

S = 0.5at²

S = maximum height

Tjme taken (t) = time of flight / 2 = 5.93/2 = 2.965 s

Hence,

S = 0.5at²

S = 2 × 0.5 × 9.8 × 2.965²

S = 43.0770025

S = 43.01 m

i think it because equal matrices have equal dimensions, only square matrices can be symmetric. for all indices and . every square diagonal matrix is symmetric, since all off-diagonal elements are zero. in linear algebra, a real symmetric matrix represents a self-adjoint operator over a real inner product space.

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s