Projectile Motion without Initial velocity

[Disoo]

Homework Statement

A stone is fired from a slingshot at an angle of 65° from the horizontal. The stone strikes the ground 8.0s later at an altitude 30m lower than the height at which it was released.
At what initial velocity was the stone released?

Homework Equations

dy = viyt + 1/2gt2
Viy = Vi sin θ

The Attempt at a Solution

dy = viyt + 1/2gt2
dy = Vi sin θt + 1/2gt2
dy + 30m = Vi sin (65)(8.0s) + 1/2(+9.91m/s2)(8.0s)2
dy =7.25 Vi + 283.92 m

I'm stuck after this stage. Currently I don't see a clear solution. How could I approach the problem?

 

[rl.bhat]

Hi Disoo, welcome to PF.
In your equation you have not used the proper sign convention.
For the projectile motion the equation can be written as
y = yo + vi*sinθ*t - 1/2*g*t^2.
Take y = 0 at ground level.

 

[tomcue]

I would first check the assumptions and limitations of the given problem. In this case, the problem assumes that the stone is fired from a slingshot at an angle of 65° from the horizontal and that the only force acting on the stone is gravity. It also assumes that the stone has no initial velocity.

Next, I would use the given equations and information to analyze the problem and see if there are any other variables that can be determined. For example, we know the time of flight (8.0s) and the change in altitude (30m) but we do not know the initial velocity.

One approach to solving this problem would be to use the equation for horizontal displacement, dx = vixt, where dx is the horizontal distance traveled, vix is the initial horizontal velocity, and t is the time of flight. Since the stone lands at the same horizontal position as it was released, we can set dx equal to 0. This gives us 0 = vix(8.0s). Since we know that the stone has no initial horizontal velocity, we can conclude that vix = 0.

Another approach would be to use the equation for vertical displacement, dy = viyt + 1/2gt², where dy is the vertical displacement, viy is the initial vertical velocity, g is the acceleration due to gravity, and t is the time of flight. In this case, we know the vertical displacement (30m), the time of flight (8.0s), and the acceleration due to gravity (9.81m/s²). We can rearrange this equation to solve for the initial vertical velocity, viy. This would give us viy = -1.84m/s.

However, this approach assumes that the stone is fired from ground level. If the stone is released from a height above the ground, we would need to take into account the initial height in our calculations. In this case, we would use the equation dy = viyt + 1/2gt² + hi, where hi is the initial height. Since we know that the stone is released at a height above the ground, we can use this equation to solve for the initial velocity.

In conclusion, there are multiple ways to approach this problem and the specific approach would depend on the assumptions and limitations of the given problem. I would carefully analyze the given information and use the appropriate equations and methods to find