physics question help?

1. A child throws a ball with a velocity of 1.93 m/s [horizontally] at a height of 0.75 m. If air resistance is negligible, the horizontal range of the ball is
t = √2h/g = √1.5/9.8 = 0.391 sec
range = V*t = 1.93*0.391 = 0.7550 m


2.A rock is launched with a velocity of 12 m/s [42° above horizontal] at a height of 9.5 m above level ground. The horizontal range of the rock is
-9.5 = 12*sin42*t-4.9t^2
t = (12*0.67+√(12*0.67)^2+19.6*9.5)9.8 = 2.437 sec
x = Vcos 40*t = 12*2.437*0.74 = 21.73 m

(→) u = 1.93, s = x, t = t and a = 0

(↓) u = 0, s = 0.75 ↓, t = t and a = g => 9.8 ↓

So, using s = ut + (1/2)at² for vertical motion we have:

0.75 = 0 + (1/2)(9.8)t²

=> t² = 0.75/4.9

so, t = 0.391 seconds

Then, using s = ut + (1/2)at² for horizontal motion we get:

x = 1.93(0.391) => 0.755 metres
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(Vertical motion) u = 12sin42 (↑)...i.e. -12sin42 (↓), s = 9.5, t = t and a = g = 9.8 (↓)

(Horizontal motion) u = 12cos42, s = x, t = t and a = 0

Then, using s = ut + (1/2)at² for vertical motion we have:

9.5 = -12tsin42 + (1/2)(9.8)t²

=> 4.9t² - 12tsin42 - 9.5 = 0

Then, using the quadratic formula we get:

t = [12sin42 ± √(12sin42)² - 4(4.9)(-9.5))]/9.8

so, t = (12sin42 ± 15.833)/9.8

i.e. t = 2.43

Then, for horizontal motion we have:

x = 12(2.43)cos42 => 21.67 metres

:)>

The child was abducted by the muffin man.


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