# How to find initial velocity given distance and angle

## Homework Statement

A projectile is fired with initial velocity Vo at an angle of 20° with the horizontal. Determine the required value of Vo if the projectile is to hit point b.
point b is 3 km away.

## Homework Equations

y=(Vy)o t + 1/2 a t^2

## The Attempt at a Solution

i tried isolating time(t) from the above formula which gave me square root of 2y/a but i cant take the square root of a negative number that bring the acceleration

## Answers and Replies

Related Introductory Physics Homework Help News on Phys.org
Some hints: Draw the diagram and then use the formula. What do you know about the x-distance and y-distance?

I have to multiply it by cos20 and sin 20

Chestermiller
Mentor
Hi Tsukuba. Welcome to Physics Forums.

What is the value of y when the projectile reaches the ground again? What is the acceleration "a" in your equation for y? Can you use your equation for y to figure out how much time the projectile is in the air?

Chet

hello and thank you.

a=-9.81 m/s^2
I can use the equation because i would have 2 unknowns, that being the initial velocity and time. Like i said I tried isolating time and the formula i get is square root of 2x/a and i cant take the square root of a negative number.

Chestermiller
Mentor
hello and thank you.

a=-9.81 m/s^2
I can use the equation because i would have 2 unknowns, that being the initial velocity and time. Like i said I tried isolating time and the formula i get is square root of 2x/a and i cant take the square root of a negative number.
As I see it, your equation for the y direction is going to be:
$$0=v_0\sin(20)t-\frac{9.8}{2}t^2$$
Is this in agreement with your assessment? If so, you can solve this equation for t by factoring the right hand side, and discarding the root at t = 0. In terms of v0, what do you get for t?

Chet

instead of the 0 wouldn't i have the distance of 3km?

here is the question and the diagram

Chestermiller
Mentor
instead of the 0 wouldn't i have the distance of 3km?
We are talking about the y direction here, not the x direction. When the projectile hits the ground, y = 0.