Question

Hello,

Answer 1

Answer:

s = height above ground

s = 60 + 20 t - 4.9 t^2 (standard physics equation on earth)

at t = 0

s = 60 (clearly :)

now when does it hit the bleak earth?

That is when s = 0

4.9 t^2 - 20 t -60 = 0

solve quadratic and use the positive t (the negative t was back before you threw it if you had thrown it from the ground)

t = 6.09 or - 2.01

use t = 6.09

now to do the last part there are two obvious ways to get t at the peak

1. look for vertex of parabola

2. look for halfway between t = -2.01 and t = 6.09

I will do it the hard (11) waay by completing the square

4.9 t^2 - 20 t = -(s-60)

t^2 - 4.08 t = -.204 s + 12.2

t^2 - 4.08 t +2.04^2 = -.204 s +12.2 + 4.16

(t-2.04)^2 = -.204(s-80.2)

so

top at 80.2 meters at t = 2.04 s

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

quick check on time

should be average of 6.09 and -2.01

=4.08 /2 = 2.04 check

Answer 2

Answer:                                                                                                                 I HOPE THAT MY ANSWER WILL HELP U        

Step-by-step explanation:

s = height above ground

s = 60 + 20 t - 4.9 t^2 (standard physics equation on earth)

at t = 0

s = 60 (clearly :)

now when does it hit the bleak earth?

That is when s = 0

4.9 t^2 - 20 t -60 = 0

solve quadratic and use the positive t (the negative t was back before you threw it if you had thrown it from the ground)

t = 6.09 or - 2.01

use t = 6.09

now to do the last part there are two obvious ways to get t at the peak

1. look for vertex of parabola

2. look for halfway between t = -2.01 and t = 6.09

I will do it the hard (11) waay by completing the square

4.9 t^2 - 20 t = -(s-60)

t^2 - 4.08 t = -.204 s + 12.2

t^2 - 4.08 t +2.04^2 = -.204 s +12.2 + 4.16

(t-2.04)^2 = -.204(s-80.2)

so

top at 80.2 meters at t = 2.04 s

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

quick check on time

should be average of 6.09 and -2.01

=4.08 /2 = 2.04 check