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3 years ago

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A ball is thrown from 1 m above the ground. The initial velocity is 20 m/s at an angle of 40 degrees above the horizontal. What

is the maximum height of the ball above the ground

1 answer:

Kipish [7]3 years ago

5 0

Answer:

9.4 m

Explanation:

This is a projectile motion. The maximum height above the level projection is given by

$H = \dfrac{(v_0\sin\theta)^2}{2g}$

$v_0$ is the initial velocity, $\theta$ is the angle of projection above the horizontal and $g$ is the acceleration of gravity.

$H = \dfrac{(20\sin40^\circ)^2}{2\times9.8} = \dfrac{165.27}{19.6} = 8.4\text{ m}$

The maximujm height above the ground is 1 m + 8.4 m = 9.4 m

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An object is thrown straight up into the air with an initial speed of 8 m/s, and reaches a greatest height of 15 m before it fal

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Answer:

The value is $T_t = 2.5659 \ s$

Explanation:

From the we are told that

The initial speed of the object is $u = 8 \ m/s$

The greatest height it reached is $h = 15 \ m$

Generally from kinematic equation we have that

$v^2 = u^2 + 2gH$

At maximum height v = 0 m/s

So

$0^2 = 8^2 + 2 * - 9.8 * H$

=> $H = 3.27 \ m$

Here H is the height from the initial height to the maximum height

So the initial height is mathematically represented as

$s = h - H$

=> $s = 15 - 3.27$

=> $s = 11.73 \ m$

Generally the time taken for the object to reach maximum height is mathematically evaluated using kinematic equation as follows

$v = u + (-g) t$

At maximum height v = 0 m/s

$0 = 8 - 9.8t$

=> $t = 0.8163 \ s$

Generally the time taken for the object to move from the maximum height to the ground is mathematically using kinematic equation as follows

$h = ut_1 + \frac{1}{2} gt_1^2$

Here the initial velocity is 0 m/s given that its the velocity at maximum height

Also g is positive because we are moving in the direction of gravity

So

$15 = 0* t + 4.9 t^2$

=> $t_1 = 1.7496$

Generally the total time taken is mathematically represented as

$T_t = t + t_1$

=> $T_t = 0.8163 + 1.7496$

=> $T_t = 2.5659 \ s$

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If a high jumper needs to make his center of gravity rise 1.50 m, how fast must he be able to sprint? Assume all of his kinetic

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Answer:

v = 5.42 m/s

Explanation:

given,

height of the jumper = 1.5 m

velocity of sprinter = ?

kinetic energy can be transformed into potential energy

$m g h = \dfrac{1}{2}mv^2$

$g h = \dfrac{1}{2}v^2$

$v =\sqrt{2gh}$

$v =\sqrt{2\times 9.8 \times 1.5}$

v = 5.42 m/s

Speed of the sprinter is equal to v = 5.42 m/s

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3 years ago

The action of two forces. One is a forward force of 1157 N provided by traction between the wheels and the road. The other is a

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Complete question

A 2700 kg car accelerates from rest under the action of two forces. one is a forward force of 1157 newtons provided by traction between the wheels and the road. the other is a 902 newton resistive force due to various frictional forces. how far must the car travel for its speed to reach 3.6 meters per second? answer in units of meters.

Answer:

The car must travel 68.94 meters.

Explanation:

First, we are going to find the acceleration of the car using Newton's second Law:

$\sum\overrightarrow{F}=m\overrightarrow{a}$ (1)

with m the mass , a the acceleration and $\sum\overrightarrow{F}$ the net force forces that is:

$(F-f)$ (2)

with F the force provided by traction and f the resistive force:

(2) on (1):

$(F-f)=ma$

solving for a:

$a=\frac{F-f}{m} =\frac{1157N-902N}{2700kg} =0.094\frac{m}{s^{2}}$

Now let's use the Galileo’s kinematic equation

$Vf^{2}=Vo^{2}+2a\varDelta x$ (3)

With Vo te initial velocity that's zero because it started from rest, Vf the final velocity (3.6) and $\varDelta x$ the time took to achieve that velocity, solving (3) for $\varDelta x$ :

$\varDelta x= \frac{Vf^{2}}{2a} = t= \frac{(3.6\frac{m}{s})^2}{2*0.094\frac{m}{s^{2}}}$

$t=68.94 m$

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3 years ago