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bagirrra123 [75]
3 years ago
13

A golf ball is projected upward from ground level at an initial velocity of 112 ft/sec. The height of a projectile can be modele

d by s(t) = – 16t^2 + v_0t + s_0, where t is time in seconds, s_0 is the initial height in feet, and v_0 is the initial velocity in ft/sec.

Physics
1 answer:
777dan777 [17]3 years ago
5 0

Answer:

The ball reaches a maximum height of 196ft above the point where it was launched.

Explanation:

Please see attachment below.

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Eduardwww [97]

Multiply the mass by the gravity.

1200kg x -9.8= -11760 N

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A car traveling at 27 m/s slams on its brakes to come to a stop. It decelerates at a rate of 8 m/s2 . What is the stopping dista
qaws [65]

<em>v</em>² - <em>u</em>² = 2 <em>a</em> ∆<em>x</em>

where <em>u</em> = initial velocity (27 m/s), <em>v</em> = final velocity (0), <em>a</em> = acceleration (-8 m/s², taken to be negative because we take direction of movement to be positive), and ∆<em>x</em> = stopping distance.

So

0² - (27 m/s)² = 2 (-8 m/s²) ∆<em>x</em>

∆<em>x</em> = (27 m/s)² / (16 m/s²)

∆<em>x</em> ≈ 45.6 m

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What erases the impact craters on earth and is responsible for most of the landforms that we see?
frez [133]
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John ( body mass= 65 kg) is taking off for a long jump . Horizontal accerleration ax is 5m/s^2 and vertical acceleration ay is 0
cestrela7 [59]

Answer:

(a) The horizontal ground reaction force  F_{g,x}=325\, N

(b)  The vertical ground reaction force F_{g,y}=696\, N

(c)   The resultant ground reaction force F_g=768\, N

Explanation:  

Given

John mass , m = 65 kg

Horizontal acceleration , a_x= 5.0 \frac{m}{s^{2}}

Vertical acceleration , a_y=0.9 \frac{m}{s^{2}}

(a) Using Newton's 2nd law in horizontal direction

F_{g,x}=ma_x

=>F_{g,x}=65\times 5\, N=325\, N

Thus the horizontal ground reaction force  F_{g,x}=325\, N

(b) Using Newton's 2nd law in vertical direction

F_{g,y}-mg=ma_y

=>F_{g,y}=mg+ma_y

=>F_{g,y}=65\times (9.81+0.9)\, N=696\, N

Thus the vertical ground reaction force F_{g,y}=696\, N

(c)  Resultant ground reaction force is

F_g=(F_{g,x}^{2}+F_{g,y}^{2})^{\frac{1}{2}}

=>F_g=(325^{2}+696^{2})^{\frac{1}{2}}\, N=768\, N

=>F_g=768\, N

Thus  the resultant ground reaction force F_g=768\, N

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