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Zolol [24]
3 years ago
12

A baseball is hit with a speed of 27.0 m/s at an angle of 47.0 ∘ . It lands on the flat roof of a 10.0 m -tall nearby building.

Part A If the ball was hit when it was 1.3 m above the ground, what horizontal distance does it travel before it lands on the building?
Physics
1 answer:
fenix001 [56]3 years ago
3 0

Answer:

H = Vy t - 1,2 g t^2            formula for height of ball after t sec

H = 10 - 1.3 = 8.7 m

Vy = 27 sin 47 = 19.7 m/s     vertical speed of ball

8.7 = 19.7 t - 9.8/2 t^2       height of ball after t sec

4.9 t^2 - 19.7 + 8.7 = 0       rearranging

[19.7 ± (388 - 170)^1/2] / 2 *4.9 = [19.7 ± 14.7] / 9.8 = .51 ,3.5 sec

.51 sec would be on the way up and 3.5 sec on the way down

Sx = 27 * cos 47 * 3.5 = 64.4 m around 200 ft seems reasonable

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fredd [130]

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6 0
3 years ago
A particle with mass 1.81*10^-3kg and a charge of 1.22*10^-8C has, at a given instant, a velocity v= (3.0*10^4 m/s)j.
charle [14.2K]

Answer:

a = -0.33 m/s² k^

Direction: negative

Explanation:

From Newton's law of motion, we know that;

F = ma

Now, from magnetic fields, we know that;. F = qVB

Thus;

ma = qVB

Where;

m is mass

a is acceleration

q is charge

V is velocity

B is magnetic field

We are given;

m = 1.81 × 10^(−3) kg

q = 1.22 × 10 ^(−8) C

V = (3.00 × 10⁴ m/s) ȷ^.

B = (1.63T) ı^ + (0.980T) ȷ^

Thus, since we are looking for acceleration, from, ma = qVB; let's make a the subject;

a = qVB/m

a = [(1.22 × 10 ^(−8)) × (3.00 × 10⁴)ȷ^ × ((1.63T) ı^ + (0.980T) ȷ^)]/(1.81 × 10^(−3))

From vector multiplication, ȷ^ × ȷ^ = 0 and ȷ^ × i^ = -k^

Thus;

a = -0.33 m/s² k^

7 0
3 years ago
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