All categories

3 years ago

A 74.1 kg high jumper leaves the ground with

Physics

1 answer:

pishuonlain [190]3 years ago

7 0

Answer:

4.80 m

Explanation:

We are given the mass of the high jumper, its initial velocity, and the acceleration of gravity. We are trying to find the vertical displacement of the high jumper.

Let's set the upwards direction to be positive and the downwards direction to be negative.

List out the relevant known variables.

v₀ = 9.7 m/s
a = -9.8 m/s²
Δx = ?

We still need one more variable in order to use the constant acceleration equations. Since we are trying to find the max height of the jumper, we can use the fact that at the top of its trajectory, its final velocity will be 0 m/s.

4. v = 0 m/s

Using these four variables, let's find the constant acceleration equation that contains these variables:

v² = v₀² + 2aΔx

Substitute the known values into the equation and solve for Δx.

(0)² = (9.7)² + 2(-9.8)Δx
0 = 94.09 + (-19.6)Δx
-94.09 = -19.6Δx
Δx = 4.80

The high jumper can jump to a max height of 4.80 m.

You might be interested in

(I) A novice skier, starting from rest, slides down an icy frictionless 8.0° incline whose vertical height is 105 m. How fast is

Vlad1618 [11]

Answer:

v = 45.37 m/s

Explanation:

Given,

angle of inclination = 8.0°

Vertical height, H = 105 m

Initial K.E. = 0 J

Initial P.E. = m g H

Final PE = 0 J

Final KE = $\dfrac{1}{2}mv^2$

Using Conservation of energy

$KE_i + PE_i + KE_f + PE_f$

$0 + m g H = \dfrac{1}{2}mv^2 + 0$

$v = \sqrt{2gH}$

$v = \sqrt{2\times 9.8 \times 105}$

v = 45.37 m/s

Hence, speed of the skier at the bottom is equal to v = 45.37 m/s

3 0

3 years ago

Lee skated 36 miles in 3 hours. What was the speed at which Lee

Zanzabum

Answer:

12mph

Explanation:

36/3=mph

8 0

3 years ago

Find the range of a projectile launched at an angle of 30° with an initial velocity of 20m/s.

Tems11 [23]

Answer:

<em>The range is 35.35 m</em>

Explanation:

<u>Projectile Motion</u>

It's the type of motion that experiences an object projected near the Earth's surface and moves along a curved path exclusively under the action of gravity.

Being vo the initial speed of the object, θ the initial launch angle, and $g=9.8m/s^2$ the acceleration of gravity, then the maximum horizontal distance traveled by the object (also called Range) is:

$\displaystyle d={\frac {v_o^{2}\sin(2\theta )}{g}}$

The projectile was launched at an angle of θ=30° with an initial speed vo=20 m/s. Calculating the range:

$\displaystyle d={\frac {20^{2}\sin(2\cdot 30^\circ )}{9.8}}$

$\displaystyle d={\frac {400\sin(60^\circ )}{9.8}}$

$d=35.35\ m$

The range is 35.35 m

7 0

3 years ago

Which energy transformation occurs in the core of a nuclear reactor?

romanna [79]

The correct answer among the choices given is option B. The energy transformation that occurs in the core of a nuclear reactor is from nuclear energy to thermal energy. In a power plant nuclear fission which involves nuclear energy to heat up water around it. This part is the core of the process.

5 0

3 years ago

(a) A light-rail commuter train accelerates at a rate of 1.35 m/s2 . How long does it take to reach its top speed of 80.0 km/h,

Mashcka [7]

Answer:

(a) Time t = 16.46 sec

(b) Time t =13.466 sec