A student throws a baseball vertically upward, and 2.8s later catches it at the same level. calculate velocity

Question

neglecting air resistance.

calculate velocity at which the ball left the students hand.

Answer 1

The time to apogee is 1/2 the total flight time.

v(t)=v0-g*t
at apogee, v(t)=0
so
1.4*g=v0
using 9.8
13.7 m/s


j

Answer 2

distance up = distance down time up + time down = total time = 2.8s The baseball decelerates going up as quickly because it speeds up happening. So, t up = t down = 2.8s / 2 = a million.4s the comparable as above individual wrote yet he made it long with the aid of using distance - time relation then distance-speed relation .we are in a position to quickly use speed-time relation right here ( v = u + a*t) speed at whch the ball left the pupil = speed of the ball whilst it comes right down to the comparable point.(speed up = speed down) thinking downward action v = u + gt u = 0 ( commencing from, stopped at max top) ; g = 9.8m/s^2 ; t = a million.4 s v = 0 + 9.8 * a million.4 = 13.seventy two m sorry i'm not english speaker so i copied his phrases.

Answer 3

Use the formula v=v(intial) + at

v=0, v(intial) is what you're solving for, a=9.8 (positive or negative depending on the teacher, I use negative), t= 1.4 (you only use half because that's from beginning to end)

I got 13.72 m/s