Question
A ball is thrown horizontally from the top of a cliff 50m high with a velocity of 7m/s. How far from the base of the building does the ball land.
This is what I think is how you do the question(Please correct me if I am wrong).
y=vt+.5a(t)²
50=0+.5(9.8)t²
10.204=t²
t=3.194 seconds
x=vt+.5a(t)²
x=(7)(3.194)+.5(0)(3.194)²
x=22.36 m
So is the answer 22.36m?
Please help me if I go wrong
This is what I think is how you do the question(Please correct me if I am wrong).
y=vt+.5a(t)²
50=0+.5(9.8)t²
10.204=t²
t=3.194 seconds
x=vt+.5a(t)²
x=(7)(3.194)+.5(0)(3.194)²
x=22.36 m
So is the answer 22.36m?
Please help me if I go wrong
Answers
bobpursley
time is right. Howver, on the horizontal equation, a=0
x=vt
x=vt
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