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# i need help with these physics problem?

1. A cannonball shot with an initial velocity of 147 m/s at an angle of 46° follows a parabolic path and hits a balloon at the top of its trajectory. Neglecting air resistance, how fast is the cannonball going when it hits the balloon?

m/s

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2. Students in lab measure the speed of a steel ball launched horizontally from a table top to be 4.3 m/s. If the table top is 1.1 m above the floor, where should they place a 20 cm tall tin coffee can to catch the ball when it lands?

m

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3. John and Tracy look from their 80 m high-rise balcony to a swimming pool below - not exactly below, but rather 24 m out from the bottom of their building. They wonder how fast they would have to jump horizontally to succeed in reaching the pool. What is the answer?

m/s

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### 1 Answer

- 1 decade agoFavorite Answer
I use g = 10 m/s^2 for all three, so you have to be careful if you use g as 9.81 m/s^2

For (1),

At the top of the trajectory, the vertical component of velocity is 0.

i.e. V(y) at top = 0,

Therefore, the velocity of the cannonball at the top is the horizontal component of velocity, V(x),

V(x) = V cos Q (I use Q as Theta)

= 147 x cos 46

= 102.12 m/s (answer)

For (2),

V(x) = 4.3 m/s and it doesn't change throughout the path,

since the height of the can is 20 cm(0.2 m) ;

using s = ut + 0.5at^2

s(y) = 0.5 x 10 x t^2 (since u(y) = 0 b4 the ball drops)

(1.1-0.2) = 0.5 x 10 x t^2

t^2 = 0.18

t = 0.4243 s

s(x) = u(x)t (since horizontal component doesn't undergo acceleration)

s(x) = 4.3 x 0.4243

= 1.8245 m from the table (answer)

For (3),

using s = ut + 0.5at^2

s(y) = uy(t) + 0.5gt^2

80 = 0.5 x 10 x t^2 (since u(y) = 0 before they jump)

t^2 = 16

t = 4 s

s(x) = u(x) x t

24 = u(x) x 4

u(x) = 6 m/s (answer)