In downtown, children bring their ball to the top of a tower and compete for accuracy in hitting a target on the ground. Suppose that the tower is 9m high and that the bull's eye is a horizontal distance of 3.5m from the launch point. (a) if the ball is thrown horizontally, what is the launch speed needed to hit the bull's eye? (B) in the previous problem the ball is given an initial horizontal speed 3.3m/s, what are the magnitude and direction after 0.75 seconds and before it lands?

a. 2.58 m/s b. at t = 0.75 speed is 8.05 m/s at -65.8°, just before it lands, speed is 13.68 m/s at -76°

Explanation:

a. We first calculate its final vertical speed, v using v² = u² + 2as where u = initial vertical speed = 0, a = -g = -9.8 m/s² and s = tower height = -9 m. (negative height)

So, v² = u² + 2as = 0² + 2 × -9.8 × -9 = 176.4 ⇒ v = √176.4 = 13.28 m/s

we now use v = u + at to calculate the time it takes the ball to reach the bull's eye

v = u + at, v = -13.28 m/s (vertically downwards)

-13.28 = 0 + -9.8t = 9.8t ⇒ t = 13.28/9.8 = 1.36 s

We now use d/t = v to calculate the initial horizontal speed. Here d = horizontal distance = 3.5 m and t = 1.36 s

v = d/t = 3.5/1.36 = 2.58 m/s

b. since its initial horizontal speed is now u = 3.3 m/s. Using v₁ = u₁ + at to calculate its vertical speed at t = 0.75 s, u₁ = initial vertical speed = 0 m/s, a = -g = -9.8 m/s²

So, v = u₁ + at = 0 -9.8 × 0.75 = -7.35 m/s

So the speed at t = 0.75 s is v = √v₁² + u² = √[(-7.35)² + 3.3²] = √64.9125 = 8.05 m/s. Its direction θ = tan⁻¹(-7.35/3.3) = -65.82° ≅ -65.8°

Before it lands, its vertical velocity is -13.28 m/s. So its speed is v = √[(-13.28)² + 3.3²] = √187.2484 = 13.68 m/s. Its direction is θ = tan⁻¹(-13.28/3.3) = -76.04° ≅ -76°

b.

explanation:

hi,

what occurs during a solar eclipse?

there are small tides across earth.

the moon is between earth and the sun.

p.s. im not sure if the first answer is correct because during a solar ecliphse there are spring tides and i dont know if they are the same.