A ball is thrown in the air. Its motion is described by x = 11.0(m/s) t, y = 1.2

Question

A ball is thrown in the air. Its motion is described by x = 11.0(m/s) t, y = 1.2m + 12.0(m/s) t - 4.90 (m/s^2) t^2 Find A.) the initial position and velocity of the ball when itis thrown (at t = 0s) B.) its acceleration C.) the time when it reaches the highest point D.) the horizontal distance when it hits the ground(y=0). A ball is thrown in the air. Its motion is described by x = 11.0(m/s) t, y = 1.2m + 12.0(m/s) t - 4.90 (m/s^2) t^2 Find A.) the initial position and velocity of the ball when itis thrown (at t = 0s) B.) its acceleration C.) the time when it reaches the highest point D.) the horizontal distance when it hits the ground(y=0).

Explanation / Answer

A ball is thrown in the air. Its motion is described by x = 11.0(m/s) t, y = 1.2m + 12.0(m/s) t - 4.90 (m/s^2) t^2 Find A.) the initial position and velocity of the ball when itis thrown (at t = 0s)
x=0 and y=1.2m
so initial position=y=1.2m
velocity=vx=dx/dt=11m/sec
vy=12-9.8t=12m
so v=1.44+144=12.05m/sec

B.) its acceleration
acceleration=-9.8m/sec2
C.) the time when it reaches the highest point
u=vy=12-9.8t
at the highest point vy=0
a=-9.8m/sec2
using v=u-gt=>0=12-9.8t-9.8t=>t=0.6122sec

D.) the horizontal distance when it hits the ground (y=0).
time=2*0.6122=1.224sec

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