Projectile Motion, speed of ball

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SUMMARY

The discussion focuses on calculating the speed of a ball shot at a goal from a horizontal distance of 5.3 meters at an angle of 48 degrees below the goal height of 1.2 meters. The key equations provided for determining the components of speed are Vx = V * cos(θ) and Vy = V * sin(θ). The horizontal and vertical motion equations are x(t) = Vx * t and y(t) = -gt²/2 + Vy * t + Y, respectively. These equations allow for the calculation of the required speed to ensure the ball reaches the goal.

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Lara
I can't seem to get the right answer to this question so if someone could guide me through, please?

A ball is shot at a goal from a horizontal distance of 5.3m. The ball is released at an angle of 48 degrees to the horizontal and 1.2m below the height of the goal. What is the ballls speed?

thankyou
 
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Did you give us all the information you have? I see no way, with the given information, to specify a single speed. Do you want the speed required to hit the goal?
 
Assuming that you only need the projectile to be at the goal (5.3,1.2) you can get the speed with the following equations.

1. The components of the speed are Vx=VCosΘ and
Vy=VSinΘ

V is the speed, Θ=48deg

2. x(t) = Vxt with the condition that x(T)=5.3

3. y(t)= -gt2/2 + Vyt + Y with conditions y(0)=0 and Y(T)=1.2

This should be a nice start for you.
 

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