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2 years ago

A bullet fired vertically at a velocity of 36m/s .after 45 the bullet hit the top of a bulid how height is a bulid?

Physics

1 answer:

zzz [600]2 years ago

3 0

Answer:

The height of the building is 8,302.5 m

Explanation:

Given;

velocity of the projectile, u = 36 m/s

time of motion, t = 45 s

Let the upward direction of the bullet be negative,

The height of the building is calculated as;

$h = ut - \frac{1}{2} gt^2\\\\h = (36\times 45) - (\frac{1}{2} \times 9.8 \times 45^2)\\\\h = 1620 - 9922.5\\\\h = -8,302.5 \ m\\\\The \ height \ of \ the \ building \ is \ 8,302.5 \ m$

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A firecracker in a coconut blows the coconut into three pieces. two pieces of equal mass fly off south and west, perpendicular t

netineya [11]

Here we have mass that moves at ceratin speed. This means that we have momentum. The law that must be observed is law of conservation of momentum. It states that momentum before certain event is equal to a momentum after that event. Here we have three masses so we can write this as:
$m_{1} v_{1i} + m_{2} v_{2i} + m_{3} v_{3i} = m_{1} v_{1f} + m_{2} v_{2f} + m_{3} v_{3f}$
Before the firecracker blows a coconut does not move, so left side is equal to 0:
$0 = m_{1} v_{1f} + m_{2} v_{2f} + m_{3} v_{3f}$

We know that m1=m2=m and m3=2m. Also we are asked to find v3f so we can rewrite formula:
$v_{3f} = - \frac{m_{1} v_{1f} + m_{2} v_{2f} }{ m_{3} }$

We must take in consideration that two parts with same mass do not move in same direction. The center of mass of these two parts moves between them at angle of 45° with respect to both south and west. The speed of a center of mass is:
$v_{f} = \sqrt{ v_{1f}^{2}+ v_{2f}^{2} } \\ \\ v_{f} = 33.9m/s$
This speed we can insert into formula for v3f:
$v_{3f} = - \frac{m*33.9+m*33.9 }{ 2m } \\ \\ v_{3f} = - \frac{2m*33.9 }{ 2m } \\ \\ v_{3f} = - 33.9m/s$

We can see that part of a coconut with biggest mass has same speed as center of mass of two other parts. Negative sign shows that direction is opposite to direction of two pats. Biggest part moves towards north-east.

8 0

3 years ago

Read 2 more answers

In one cycle a heat engine absorbs 450 J from a high-temperature reservoir and expels 290 J to a low-temperature reservoir. If t

Vitek1552 [10]

Answer:

So the ratio will be $\frac{T_L}{T_H}=-0.171$

Explanation:

We have given heat engine absorbs 450 joule from high temperature reservoir

So $Q=450j$

As the heat engine expels 290 j

So work done W = 290 J

We know that efficiency $\eta =\frac{W}{Q}=\frac{290}{450}=0.6444$

It is given that efficiency of the engine only 55 % of Carnot engine

So efficiency of Carnot engine $=\frac{0.6444}{0.55}=1.171$

Efficiency of Carnot engine is $\eta =1-\frac{T_L}{T_H}$

$1.171 =1-\frac{T_L}{T_H}$

$\frac{T_L}{T_H}=-0.171$

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3 years ago

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If an object's mass increases, its------------------------- can be increased

otez555 [7]

Answer:

both kinetic energy and potential energy

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2 years ago

A ball is thrown horizontally from the top of a building 37.5 m high. The ball strikes the ground at a point 80.3 m from the bas

trasher [3.6K]

Answer:

t = 2.77 s

Explanation:

The ball in its movement describes a curved line called a semiparabola, therefore two coordinates are required to fix the position at each instant of time, since the movement is performed in the X-Y plane:

Equation of movement of the ball in the X axis

X = v₀x*t Equation (1)

Equation of movement of the ball in the Y axis

Y = y₀+ v₀y*t -½ g*t² Equation (2)

Where

X : horizontal position in meters (m)

Y : vertical position in meters (m)

y₀ : initial vertical position in meters (m)

v₀x : X-initial speed in m/s

v₀y : Y-initial speed in m/s

g: acceleration due to gravity in m/s²

t : time to position (X,Y)

Data

y₀ = 37.5 m

v₀y = 0

g = 9.8 m/s²

Problem development

The time the ball remains in the air is the same as the ball takes to touch the floor, that is, Y = 0

We apply the Equation (2):

Y = y₀+ (v₀y)*t - (½) g*t²

0 = 37.5 +(0)*t- (1/2)*(g)*t²

0 = 37.5 - (1/2)*(9.8)*t²

(1/2)*(9.8)*t² = 37.5

t² = (2)(37.5)/(9.8)

$t= \sqrt{\frac{2*37.5}{9.8} }$

t = 2.77 s

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2 years ago

For years, the tallest tower in the United States was the Phoenix Shot Tower in Baltimore, Maryland. The shot tower was used fro

Mariulka [41]

Answer:

The velocity of the droplet right before it hits the ground is 40.08 m/s.

Explanation:

To determine the velocity of the droplet right before it hits the ground,

From one of the equations of kinematic for free fall motions,

v = u + gt

Where v is the final velocity

u is the initial velocity

g is acceleration due to gravity (take g = 9.8 m/s²)

and t is time

For the question, v is the velocity of the droplet right before it hits the ground.

u = 0 m/s (Since the molten lead was dropped from rest)

Therefore,

v = gt

First, we will determine the time t

Also, from one of the equations of kinematic for free fall motions,

h = ut + 1/2(gt²)

u = 0 m/s

From the question, the molten lead was dropped from the top of the 82.15 m tall tower, therefore

h = 82.15

Hence,

82.15 = 0×t + 1/2 (9.8 × t²)

82.15 = 1/2 (9.8 × t²)

82.15 = 4.9 t²

t² = 82.15/4.9

∴ t = 4.09 secs

Now, for the velocity v, of the droplet right before it hits the ground,

Recall

v = gt

Then,

v = 9.8 × 4.09

v = 40.08 m/s

Hence, the velocity of the droplet right before it hits the ground is 40.08 m/s.

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3 years ago

A bullet fired vertically at a velocity of 36m/s .after 45 the bullet hit the top of a bulid how height is a bulid?​

A bullet fired vertically at a velocity of 36m/s .after 45 the bullet hit the top of a bulid how height is a bulid?