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.
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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.
To find the maximum height reached above the ground, we can use the equations of motion. We know that the initial velocity (u) is 98 m/s and the displacement (s) is the height of the building, which is 100 m. The acceleration (a) due to gravity is -9.8 m/s^2 (negative because it acts in the opposite direction to the initial velocity).
a) To find the maximum height (H), we can use the following kinematic equation:
v^2 = u^2 + 2as
Where v is the final velocity at the maximum height and s is the displacement. At maximum height, the final velocity (v) is 0 since the bullet momentarily stops.
0 = (98 m/s)^2 + 2*(-9.8 m/s^2) * H
Simplifying this equation, we get:
0 = 9604 - 19.6H
Rearranging the equation to isolate H, we have:
H = 9604 / 19.6
Calculating this, we find that the maximum height reached above the ground is approximately 490.82 meters.
b) To find the total time before reaching the ground, we can use the following equation:
v = u + at
Where v is the final velocity at the ground (which is negative since it's in the opposite direction to the initial velocity) and t is the total time.
0 = 98 m/s + (-9.8 m/s^2) * t
Rearranging the equation gives:
9.8t = 98
t = 98 / 9.8
Calculating this, we find that the total time before reaching the ground is approximately 10 seconds.
c) Finally, to find the velocity upon landing, we can use the equation mentioned earlier:
v = u + at
where v is the final velocity at the ground, u is the initial velocity, a is the acceleration due to gravity, and t is the total time calculated in part b.
Plugging in the values, we get:
v = 98 m/s + (-9.8 m/s^2) * 10 s
Simplifying this equation, we find:
v = 98 m/s - 98 m/s
v = 0 m/s
Thus, the velocity upon landing is 0 m/s.
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.