Question

Question #75e5d

Answer 1

The answer is d) 7 seconds.

Explanation:

Because the ball was launched at a `30^@` angle with the horizontal, its initial velocity will have a vertical component and horizontal component.

`{(v_(0x) = v_0 * cos(30^@)), (v_(0y) = v_0 * sin(30^@)) :}`

Since you have no information about the distance from the tower the ball travelled in its flight, you're going to have to turn to the vertical component of its trajectory.

So, vertically, the ball starts at a height of 70 m. This means that, when it reaches ground level, the total vertical displacement will be equal to -70 m.

Think of its like this - when it hits the ground, the ball is 70 meters lower than its starting position.

This means that you can write

`d_"vert" = v_(0y) * t - 1/2 * g * t^2`, where

`d_"vert"` - the total vertical displacement;
`v_(0y)` - the vertical component of its initial velocity;
`t` - the total time of flight.

Plug in your values into the above equation to get (use `g=10"m"/"s"^2`)

`-70cancel("m") = 50cancel("m")/cancel("s") * underbrace(sin(30^@))_(color(blue)(=1/2)) * t - 1/2 * 10 cancel("m")/cancel("s"^2) * t^2`

`-70 = 25 * t - 5t^2 <=> -5t^2 + 25t +70 = 0`

Solving this quadratic equation will get you

`t_(1,2) = (-25 +- sqrt(25^2 - 4 * (-5) * 70))/(2 * (-5))`

`t_(1,2) = (-25 +- 45)/(-10) => {(cancel(t_1 = 20/-10 = -2)), (t_2 = -70/-10 = 7) :}`

Therefore, the ball reaches the ground in

`t_"flight" = color(green)("7 s")`