Question

A dart is thrown horizontally at a target's center that is 5.00 m 5.00m5, point, 00, start text, m, end text away. The dart hits the target 0.150 m 0.150m0, point, 150, start text, m, end text below the target's center. We can ignore air resistance.

Answer 1

The question is incomplete and doesn't have any requirement. I'm computing the angle at which the dart should have been thrown to hit in the center of the target

Answer:

The dart should be thown with an angle of elevation of $\displaystyle \theta =1.72^o$

Explanation:

Horizontal and Diagonal Motion

When an object is launched horizontally, its initial vertical speed is zero and the horizontal component remains the same in time. If the initial velocity forms an angle with the horizontal, then the vertical velocity has an initial value which can be positive or negative depending on the angle (above or below the horizontal).

The dart is thrown horizontally and when it reaches the target, 5 meters away, the center of the target was missed by 0.15 meters below. The question doesn't have any requirement, but I'll assume you want to know at what angle should the dart be thrown at the same speed, to be exact on target.

Let's start with the original horizontal launching. The distance x is computed as

$\displaystyle x=v_o.t$

The distance y is

$\displaystyle y=\frac{g\ t^2}{2}$

We need to find the value of $v_o$, given x and y. Solving for t in the last equation

$\displaystyle t=\sqrt{\frac{2y}{g}}$

Replacing in x

$\displaystyle x=v_o\sqrt{\frac{2y}{g}}$

Solving for $v_o$

$\displaystyle v_o=x\sqrt{\frac{g}{2y}}$

Now, let's suppose we try to reach the center of the target by throwing the dart at the same speed but with a certain angle $\theta$ above the horizontal. That way, the formulas for x changes to

$\displaystyle x=\frac{v_o^2\ sin2\theta}{g}$

Replacing the value of $v_o$

$\displaystyle x=\left(x\sqrt{\frac{g}{2y}}\right)^2\ \frac{\ sin2\theta}{g}$

Operating

$\displaystyle x=\frac{x^2g}{2y}\ \frac{sin2\theta }{g}$

Simplifying

$\displaystyle 2y=x\ sin2\theta$

Solving for the angle

$\displaystyle \ sin2\theta= \frac{2y}{x}$

$\displaystyle \ sin2\theta= \frac{2(0.15)}{5}=0.06$

$\displaystyle 2\theta =3.44^o$

$\displaystyle \theta =1.72^o$

Answer 2

Answer:

28.6 m/s

Explanation:

the verified expert clearly isnt an expert no shade tho