3. MOTION IN SPACE
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Problem (Page 786 #51(c)). A baseball is hit from a height of 3 feet
with initial speed 120 feet per second and at an angle of 31 degrees above the
horizontal. Find a vector-valued function describing the position of the ball t
seconds after it is hit. To be a home run, the ball must clear a wall that is 385
feet away and 6 feet tall. Determine whether this is a home run.
Solution
g ⇡ 32 ft/sec. Newton’s 2nd Law =)
v0(t) = a(t) = gj = 32j =)
Z
Z
v(t) = a(t) dt = ( 32j) dt =
mgj = F(t) = ma(t).
32tj + c1.
v(0) = h0, 0i+c1 = h120 cos 31 , 120 sin 31 i =) c1 = h120 cos 31 , 120 sin 31 i.
Thus v(t) = h120 cos 31 , 120 sin 31
32ti.
R
r(t) = v(t) dt = h120 cos 31 t, 120 sin 31 t
r(0) = h0, 0i + c2 = h0, 3i =) c2 = h0, 3i.
Thus r(t) = h120 cos 31 t, 120 sin 31 t
16t2i + c2.
16t2 + 3i.
When first component = 120(cos 31 )t = 385, t = 3.75 =)
second component = 120(sin 31 )(3.75)
16(3.75)2 + 3 = 9.77 =) a home run.