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3 years ago

Please help!!!

Physics

1 answer:

Lunna [17]3 years ago

5 0

Answer: 1. 8cm/s²

2. 9cm/s²

Explanation:

1. Use equation for acceleration: a=(V2-V1)/t

V1 Speed on start

V2 speed on finish

V1=0 m/s

V2=24cm/s=0.24m/s

t=3s

a=(0.24m/s-0m/s)/3s

a=0.24m/s/3s

a=0.08m/s²=8cm/s

----------------------------------------------

2.Use equation for acceleration: a=(V2-V1)/t

V1 Speed on start

V2 speed on finish

V1=0cm/s

V2=36cm/s

t=4s

a=(36cm/s-0cm/s)/4s

a=36cm/s/4s

a=9cm/s²

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A stone is catapulted at time t = 0, with an initial velocity of magnitude 19.9 m/s and at an angle of 39.9° above the horizonta

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Answer:

Part a)

$x = 15.76 m$

Part b)

$y = 7.94 m$

Part c)

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Part d)

$y = 7.49 m$

Part e)

$x = 83.23 m$

Part f)

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Explanation:

As we know that catapult is projected with speed 19.9 m/s

so here we have

$v_x = 19.9 cos39.9$

$v_x = 15.3 m/s$

similarly we have

$v_y = 19.9 sin39.9$

$v_y = 12.76 m/s$

Part a)

Horizontal displacement in 1.03 s

$x = v_x t$

$x = (15.3)(1.03)$

$x = 15.76 m$

Part b)

Vertical direction we have

$y = v_y t - \frac{1]{2}gt^2$

$y = (12.76)(1.03) - 4.9(1.03)^2$

$y = 7.94 m$

Part c)

Horizontal displacement in 1.71 s

$x = v_x t$

$x = (15.3)(1.71)$

$x = 26.16 m$

Part d)

Vertical direction we have

$y = v_y t - \frac{1]{2}gt^2$

$y = (12.76)(1.71) - 4.9(1.71)^2$

$y = 7.49 m$

Part e)

Horizontal displacement in 5.44 s

$x = v_x t$

$x = (15.3)(5.44)$

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Part f)

Vertical direction we have

$y = v_y t - \frac{1]{2}gt^2$

$y = (12.76)(5.44) - 4.9(5.44)^2$

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