Question

a ball projected vertically upwards from the top of the tower 60 meters high with a velocity of 30 meters
per second. what is the maximum height above the ground? how long does it take to reach the ground?

Answer 1

the height is modeled by the function

h(t) = 60 + 30t - 4.9t^2

max height is at the vertex of the parabola, at t = -b/2a = 30/9.8

Use that to evaluate h(t) there.

It hits the ground when the height has dropped to zero. So, just solve

60 + 30t - 4.9t^2 = 0

Answer 2

V²=u²-2gh
At hmax v=0
0=30²-2x10xh
0=900-20h
20h=900
h= 900÷ 20
h=45
hmax = 45 + 60= 105 m

Answer 3

V²=u²-2gh
At hmax v=0
0=30²-2x10xh
0=900-20h
20h=900
h= 900÷ 20
h=45
hmax = 45 + 60= 105 m

For time
V=u + gt
0=30-10t
10t=30
t=30÷10
Time taken to reach the ground
h=ut+½gt²
U=0 when body falls downward
105=0xt +½x10xt²
105=5t²
t²= 21
t=4.5
Total time taken to reach the ground=>3+4.5=7.5 seconds.

Answer 4

Thanks

Answer 5

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your beauty

Answer 6

Thanks

Answer 7

Beauty i love u

Answer 8

I love the way you rote it

Answer 9

Baby thank u love 💕 you

Answer 10

For time
V=u + gt
0=30-10t
10t=30
t=30÷10
Time taken to reach the ground
h=ut+½gt²
U=0 when body falls downward
105=0xt +½x10xt²
105=5t²
t²= 21
t=4.5
Total time taken to reach the ground=>3+4.5=7.5 seconds.

Answer 11

The same as there's