Question

A stone is thrown vertically upward with a velocity of 40m/s.
A). How high will it rise?
B). How long will it take to reach its highest point?
C). What is the total time that the stone is in the air?
D). What is the velocity at the end of 2 seconds?

Answer 1

hey some more kinematic equation problems!

it seems that this is the section of your homework.
use this equation on part A

V = V + at
the left V is the final velocity (meters/ second)
the right V is the starting veloicty (meters/ second
the a is for acceleration (meters / second^2)
the t is for time, (seconds)

so lets have the initial velocity be the time when the ball is thrown up, and the final velocity be when the ball reaches it's peak.

plug in what you can figure out and i'll help with the rest

Answer 2

so the equation would read 0m/s =45m/s+(-9.8m/s^2)(t)?

Answer 3

go with the formula v^2- u^2= -2gh for the first question where v is the final vel. and u is the initial vel. g is the acceleration due to gravity and h is the height.

here -g is used since it is acting against gravity.

For the second use the equation,

h=ut-1/2gt^2

here t is time, from the first equation. Since the h is obtained so in the second you can easily find the time.

For the fourth equation use the equation v=u-gt

For the third equation use again v=u-gt
and put v=0 and u=40m/s and the time which is obtained is the total uplift span.

Answer 4

@dwade i hope now you have understood it

Answer 5

hmm alright ill work it out on paper and see what i get

Answer 6

Just make sure to use -g because it is against gravity.

Answer 7

@praxer so the equation would be 0^2 - 40^2 = -2(9.8)(h) ?

Answer 8

yes

Answer 9

81.632m is the height.

Answer 10

yes lol i just finished calculating and looked up to see the same answer i got

Answer 11

so for the second part the equation is 81.63m = 40(t) - 0.5(9.8)(t)^2?