An arrow is shot at an angle 38 ◦ with the horizontal. It has a velocity of 46 m/s. How high will the arrow go?

Physics

1 answer:

Klio2033 [76]2 years ago

8 0

Answer:

40·919 m

Explanation:

Initial velocity of the arrow = 46 m/s

Angle at which it is thrown from horizontal = 38°

<h3>At the maximum height, the vertical component of velocity will be 0</h3>

Initial velocity in vertical direction = 46 × sin(38) = 28·32 m/s

From the formula

where

v is the final velocity

u is the initial velocity

a is the acceleration

s is the displacement

Considering the formula in vertical direction and taking upward direction as positive

v = 0

u = 28·32 m/s

a = - g = - 9·8 m/s²

Let s be the maximum height

- 28·32² = - 2 × 9.8 × s

⇒ s = 40·919 m

∴ The arrow will go 40·919 m high

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A slingshot is used to launch a stone horizontally from the top of a 20.0 meter cliff. The stone lands 36.0 meters away.

alisha [4.7K]

A)

It is a launch oblique, therefore the initial velocity in the vertical direction is zero. Space Hourly Equation in vertical, we have:

$S=S_{o}+v_{o}t+ \frac{at^2}{2} \\ 20= \frac{10t^2}{2} \\ t=2s$

Through Definition of Velocity, comes:

$\Delta v= \frac{\Delta S}{\Delta t} \\ v_x= \frac{36}{2} \\ \boxed {v_{x}=18m/s}$

B)

Using the Velocity Hourly Equation in vertical direction, we have:

$v_{y}=v_{y_{o}}+gt \\ v_{y}=10\times2 \\ \boxed {v_{y}=20m/s}$

The angle of impact is given by:

$cos(\theta) =\frac{v_{x}}{v_{y}} \\ cos(\theta) = \frac{18}{20} \\ cos(\theta) =0.9 \\ arccos(0.9)=\theta \\ \boxed {\theta \approx 25.84}$

If you notice any mistake in my english, please let me know, because i am not native.