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3 years ago

if you throw a rock straight up at a speed of 19m/s. How long goes it take the ball to reach its maximum height?

Physics

1 answer:

Scrat [10]3 years ago

3 0

Answer:

It requires 1.9 seconds to reach maximum height.

Explanation:

As per given question,

Initial velocity (U) =19 m/s

Final velocity (V) = 0 m/s

$\text { Taking acceleration due to gravity }(a)=10 \mathrm{m} / \mathrm{s}^{2}$

Maximum height = S

Time taken is "t"

Calculating time taken to reach maximum height:

We know that time taken to reach the maximum height is calculated by using the formula V = U + at

Substitute the given values in the above equation.

Final velocity is “0” as there is no velocity at the maximum height.

$0=19+10 \times t$

$-19=10 \times t$

$\frac{-19}{10}=t$

t = 1.9 seconds.

The time taken to reach maximum height is 1.9 seconds.

Calculating maximum height:

$\text { Consider the equation } V^{2}-U^{2}=2 a S$

Solving the equation we will get the value of S

$0-19^{2}=2 \times(-10) \times \mathrm{S} .(-\text { is due to opposite of gravity) }$

-361 = -20S

Negative sign cancel both the sides.

$\mathrm{S}=\frac{361}{20}$

S = 18.05 m

Maximum height is 18.05 m .

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If he leaves the ramp with a speed of 31.0 m/s and has a speed of 29.5 m/s at the top of his trajectory, determine his maximum h

raketka [301]

Answer:

The maximum height reached is 4.63 m.

Explanation:

Given:

Initial speed of the man (u) = 31.0 m/s

Speed at the top of trajectory ( $u_x$ ) = 29.5 m/s

Acceleration due to gravity (g) = 9.8 m/s²

When the man reaches the top of the trajectory, the vertical component of velocity becomes zero and hence only horizontal component of velocity acts on him.

Also, since there is no net force acting in the horizontal direction, the acceleration is zero in the horizontal direction from Newton's second law. Thus, the horizontal component of velocity always remains the same.

So, speed at the top of trajectory is nothing but the horizontal component of initial velocity.

Now, initial velocity can be rewritten in terms of its components as:

$u^2=u_x^2+u_y^2$

Where, $u_x\ and\ u_y$ are the initial horizontal and vertical velocities of the man.

Now, plug in the given values and simplify. This gives,

$(31.0)^2=(29.5)^2+u_y^2\\\\961=870.25+u_y^2\\\\u_y^2=961-870.25\\\\u_y^2=90.75\ m^2/s^2--------1$

Now, we know that, for a projectile motion, the maximum height is given as:

$H=\frac{u_y^2}{2g}$

Plug in the value from equation (1) and 9.8 for 'g' to solve for 'H'. This gives,

$H=\frac{90.75}{2\times 9.8}\\\\H=4.63\ m$

Therefore, the maximum height reached is 4.63 m.

3 0

3 years ago

Evaluate u+xy where U=3 X=4 Y=7

Reptile [31]

Answer:

Explanation:

Given:

U=3

X=4

Y=7

u + xy

Substitute the given values to the equation:

3 + (4)(7)

3 + 28

6 0

2 years ago

The heat capacity of object B is twice that of object A. Initially A is at 300 K and B at 450 K. They are placed in thermal cont

ivann1987 [24]

Answer:

The final temperature of both objects is 400 K

Explanation:

The quantity of heat transferred per unit mass is given by;

Q = cΔT

where;

c is the specific heat capacity

ΔT is the change in temperature

The heat transferred by the object A per unit mass is given by;

Q(A) = caΔT

where;

ca is the specific heat capacity of object A

The heat transferred by the object B per unit mass is given by;

Q(B) = cbΔT

where;

cb is the specific heat capacity of object B

The heat lost by object B is equal to heat gained by object A

Q(A) = -Q(B)

But heat capacity of object B is twice that of object A

The final temperature of the two objects is given by

$T_2 = \frac{C_aT_a + C_bT_b}{C_a + C_b}$

But heat capacity of object B is twice that of object A

$T_2 = \frac{C_aT_a + C_bT_b}{C_a + C_b} \\\\T_2 = \frac{C_aT_a + 2C_aT_b}{C_a + 2C_a}\\\\T_2 = \frac{c_a(T_a + 2T_b)}{3C_a} \\\\T_2 = \frac{T_a + 2T_b}{3}\\\\T_2 = \frac{300 + (2*450)}{3}\\\\T_2 = 400 \ K$

Therefore, the final temperature of both objects is 400 K.

4 0

2 years ago

An object travels in a circular path of radius 5.0 meters at a uniform speed of 10. m/s. What is the magnitude of the object's a

vladimir1956 [14]

I think F= mv²/r
And F=ma
So, ma = mv²/r
a = v²/r
a = 100/5
a = 20 m/s

8 0

3 years ago

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List three advantages of reflecting telescopes over refracting telescopes.

Dima020 [189]

Answer:

Three advantages of Reflecting Telescope over refracting Telescope:

1. there is no chromatic aberration in the Reflecting Telescope being mirror as an objective, while Refracting Telescope can suffer chromatic aberration being using lenses.

2. image of Reflecting Telescope is brighter due to large, polished curved mirrors than Refracting Telescope

3. Reflecting Telescope is compact and portable in size than Refracting Telescope.

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3 years ago