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KIM [24]
3 years ago
7

A boy throws a snowball straight up in the air with an initial speed of 4.50 ft/s from a position 4.00 ft above the ground. The

snowball falls straight back down in to a 6 inches of snow. The snowball feels a deceleration of 100 m/s2 upon impact with the snow bank before coming to rest. (a) When does the snowball hit the top of the snow bank? (b) How far from the ground does the snowball come a rest?
Physics
1 answer:
IgorC [24]3 years ago
8 0

Answer:

a) 0.658 seconds

b) 0.96 inches

Explanation:

v=u+at\\\Rightarrow 0=4.5-32.1\times t\\\Rightarrow \frac{-4.5}{-32.1}=t\\\Rightarrow t=0.14 \s

Time taken by the ball to reach the highest point is 0.14 seconds

s=ut+\frac{1}{2}at^2\\\Rightarrow s=4.5\times 0.14+\frac{1}{2}\times -32.1\times 0.14^2\\\Rightarrow s=0.315\ ft

The highest point reached by the snowball above its release point is 0.315 ft

Total height the snowball will fall is 4+0.315 = 4.315 ft

s=ut+\frac{1}{2}at^2\\\Rightarrow 4.315=0t+\frac{1}{2}\times 32.1\times t^2\\\Rightarrow t=\sqrt{\frac{4.315\times 2}{32.1}}\\\Rightarrow t=0.518\ s

The snowball will reach the bank at 0.14+0.518 = 0.658 seconds after it has been thrown

v=u+at\\\Rightarrow v=0+32.1\times 0.518\\\Rightarrow v=16.62\ ft/s

v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-16.62^2}{2\times -100\times 3.28}\\\Rightarrow s=0.42\ ft

The snowball goes 0.5-0.42 = 0.08 ft = 0.96 inches

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d. Newton's first law

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A firefighting crew uses a water cannon that shoots water at 25.0 m/s at a fixed angle of 53.0° above the horizontal. The fire-f
zysi [14]

Answer:

8.8 m and 52.5 m

Explanation:

The vertical component and horizontal component of water velocity leaving the hose are

v_v = vsin(\alpha) = 25sin(53^0) = 25*0.8 = 19.97 m/s

v_h = vcos(\alpha) = 25cos(53^0) = 25*0.6 = 15 m/s

Neglect air resistance, vertically speaking, gravitational acceleration g = -9.8m/s2 is the only thing that affects water motion. We can find the time t that it takes to reach the blaze 10m above ground level

s = v_vt + gt^2/2

10 = 19.97t - 9.8t^2/2

4.9t^2 - 19.97t + 10 = 0

t= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

t= \frac{19.9658877511823\pm \sqrt{(-19.9658877511823)^2 - 4*(4.9)*(10)}}{2*(4.9)}

t= \frac{19.9658877511823\pm14.24}{9.8}

t = 3.49 or t = 0.58

We have 2 solutions for t, one is 0.58 when it first reach the blaze during the 1st shoot up, the other is 3.49s when it falls down

t is also the times it takes to travel across horizontally. We can use this to compute the horizontal distance between the fire-fighters and the building

s_1 = v_ht_1 = 15*0.58 = 8.8 m

s_2 = v_ht_2 = 15*3.49 = 52.5m

8 0
3 years ago
A toy cannon launches a 46-g golf ball straight up into the air with a kinetic energy of 6.8 J. What must the
Shkiper50 [21]

Answer : The correct option is, (C) 17 m/s

Explanation :

Formula used :

K.E=\frac{1}{2}mv^2

where,

K.E = kinetic energy = 6.8 J

m = mass of object = 46 g = 0.046 kg    (1 kg = 1000 g)

v = velocity

Now put all the given values in the above formula, we get:

K.E=\frac{1}{2}mv^2

6.8J=\frac{1}{2}\times 0.046kg\times v^2

6.8kg.m^2/s^2=\frac{1}{2}\times 0.046kg\times v^2

v=17.19m/s\approx 17m/s

Therefore, the ball's velocity be as it leaves the cannon is, 17 m/s

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3 years ago
What is the order of the events for the water cycle on a typical warm day?
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B precipitation,condensation,precipitation
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