How Far Will a Thrown Rock Travel and Will It Clear a Fence?

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A rock thrown from a height of 3 feet with a horizontal velocity of 90 ft/sec and a vertical velocity of 47 ft/sec will land approximately 270 feet away if unobstructed. The calculations involve using parametric equations to determine the trajectory, specifically x(t) and y(t) equations. To find the time of flight, set y(t) to zero and solve for t. The discussion also addresses whether the rock will clear a 7'9" fence, indicating that it will based on the calculated trajectory. The focus remains on applying parametric equations to solve the problem effectively.
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If a rock is thrown from a point 3 ft above the ground with a horizontal velocity of 90ft/sec and a vertical velocity of 47 ft/sec, how far away will it land if nothing is obstructing its path? If there is a 7'9" fence in front of you, will the rock sail over the fence?

Actually, this is a problem in my calc book. I'm supposed to solve it using parametric equations but I figured its still physics...

Using parametrics: the
x(t)=x0+h0t
x(t)=90t

y(t)=(-1/2)gt2+v0t+yo
y(t)=-16t2+47t+3

Using physics:
Not sure which equation to use. I'm a little confused. I started by adding the velocity vectors, and got 101.5 ft/s in an upwards diagonal direction
I used v2=v02+2ax
v=0, v0=101.5, solved for x and got 525.625.
answer should be 270. no idea what to do...

I'd prefer to work with the parametric method but any help is welcome...Thanks
 
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y(t)=-16t^2+47t+3
Since the rock lands on the ground, take y(t) = 0. Solve for t.
 
ah thanks. got it =)
 
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