Can anyone help me with a Math/Physics Problem?

Question

A projectile is launched upward from ground level with an initial speed of 196 meters per second. Use the position equation: h=--4.9t^2+vt, where h= height, t=time in seconds and v= initial velocity.

1. When will the projectile return to the ground?

2. How high will it go?

Answer 1

h = - 4.9t^2 + vt

h = t [v - 4.9 t]

h = 0, back on ground (remember t= not zero)

so > [v - 4.9 t] =0

t = v/4.9 (sec) >>>>>>>>>>>>>>>>>>

===================

t = 196/4.9 = 40 s

==============

t = 2 v/9.8 = v/9.8 + v/9.8 = t (going) + t (coming)

--------------------------

t (going) = [1/2] (v/4.9) = v/9.8

h (max) = - 4.9 [v/9.8]^2 + v (v/9.8)

h (max) = - 4.9 [v/9.8]^2 + (v^2/9.8)

= [v^2/9.8] [1 - (4.9/9.8)]

= v^2/2*9.8

= v^2/19.6 meter

= 196*196/19.6 = 1960 m

Answer 2

Edit: correction Vi=196m/s

Looked like "198" during first post

Thank you anil for pointing out my mistake

Thumbs up!

anil's answer is correct.

0=v-g*t

0=196-g*t

t=time up=196 / 9.8 = 20 s

return to ground: 2*20.2=40 s

How high will it go:

h=-4.9t^2+vt is the same as:

-4.9*(20^2)+196*20 = 1960 meters

check:

(196 * 20) - (1 / 2) * 9.8 * (20^2) = 1960 m