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
time is right. Howver, on the horizontal equation, a=0
x=vt
Your calculation for the time it takes for the ball to fall is correct. However, the equation you used to calculate the horizontal distance traveled by the ball is incorrect. Since the ball is thrown horizontally, the only force acting on it in the horizontal direction is its initial velocity. Therefore, you can use the formula:
x = vt
where x is the horizontal distance traveled, v is the horizontal velocity (in this case, 7 m/s), and t is the time of flight (3.194 seconds).
Plugging in the values, you get:
x = (7 m/s)(3.194 s) = 22.36 m
So the correct answer is indeed 22.36 m. Well done!