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

a) velocity at which the ball left the student's hand

answer : 14m/s

b) the heigt to which the bal climbed above the students hand

answer : 9.6m


I don't understand how to slove this problem. I have wrote the answers for the following questinos.
anyone know what equation or way to solve this problem ?

Simply apply the Equations of motion:

Equations of motion - Wikipedia, the free encyclopedia

All I've given is time and the acceleration, which is 9.8m/s
How should I go about from here?

for a) you can use

for b) you can use or

if I use your equation v = u x at

I get answer of v = (0) x (9.8m/s) ( 2.8)
which is 27.44
not 14

Quote:

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Originally Posted by benbrad

if i use your equation v = u x at

i get answer of v = (0) x (9.8m/s) ( 2.8)
which is 27.44
not 14

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You're trying to find u, not v.

Do I have to find another number to solve for initial velocity ?

There's 4 unknown and I only know 2.

I'm sorry, but could you help me step by step.
I'm a visual learner, would help where I should plot each numbers.

okay, for the first part you need to recognise that when you throw an object up and it comes back down, the first half of this motion is identical to the second half (in the absence of air resistance).

So, from this information we can determine that v = -u (same speed but in opposite directions)

using this along with v = u + at we find that

-u = u + at

Now just rearrange and solve for u.