Question

A bullet is fired vertically upward with an initial velocity of 98m/s from the top of a building 100m high, find a) the maximum height reached above the ground b) the total time before reaching the ground. C) the velocity on landing.

Answer 1

a. h=ho + (-Vo^2/2g)=100 + (-98^2)/-19.6
= 590 m.

b. V = Vo + g*Tr = 0
98 - 9.8*Tr = 0
-9.8Tr = -98
Tr = 10 s. = Rise time.

h = 0.5g*t^2 = 590
4.9*t^2 = 590
t^2 = 120.4
Tf = 11 s. = Fall time.

Tr+Tf = 10 + 11 = 21 s. = Time to reach
gnd.

c. V^2 = Vo^2 + 2g*h
Vo = 0
g = 9.8 m/s^2
h = 590 m.
Solve for V.

Answer 2

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Answer 3

h=ho + (-Vo^2/2g)=100 + (-98^2)/-19.6
= 590 m.

b. V = Vo + g*Tr = 0
98 - 9.8*Tr = 0
-9.8Tr = -98
Tr = 10 s. = Rise time.

h = 0.5g*t^2 = 590
4.9*t^2 = 590
t^2 = 120.4
Tf = 11 s. = Fall time.

Tr+Tf = 10 + 11 = 21 s. = Time to reach
gnd.

c. V^2 = Vo^2 + 2g*h
Vo = 0
g = 9.8 m/s^2
h = 590 m.