The equation of ball's height is
y = h0 + (v sinθ)t - 4.9t^2
y = 28.63t - 4.9t^2
So, just solve for y = 4.1
You will get two answers, one for when the ball is going up, and another for when it lands on the green.
can somebody show me how to find the time s step by step. A golfer hits his approach shot at an angle of
θ = 46.3°,
giving the ball an initial speed of
v0 = 39.6 m/s
(see figure below). The ball lands on the elevated green,
yf = 4.1 m
above the initial position near the hole, and stops immediately.
How much time passed while the ball was in the air?
s
4 answers
steve are you mean take the quadratic of this -4.9t^2+28.63t-4.1
your syntax is strange, but yes, the quadratic formula is the best way to find the desired time.
Vo = 39.6m/s[46.3o].
Xo = 39.6*cos46.3 = 27.4 m/s.
Yo = 39.6*sin46.3 = 28.6 m/s.
Y = Yo + g*Tr = 0 @ max ht.
28.6 + (-9.8)Tr = 0,
Tr = 2.92 s. = Rise time.
Y^2 = Yo^2 + 2g*h = 0.
28.6^2 + (-19.6)*h = 0,
h = 41.7 m. Above launching level.
ho - 0.5g*Tf^2 = 4.1.
41.7 - 4.9Tf^2 = 4.1,
Tf = 2.77 s. = Fall Time.
T = Tr+Tf = 2.92 + 2.77 = 5.69 s. = Time in air.
Xo = 39.6*cos46.3 = 27.4 m/s.
Yo = 39.6*sin46.3 = 28.6 m/s.
Y = Yo + g*Tr = 0 @ max ht.
28.6 + (-9.8)Tr = 0,
Tr = 2.92 s. = Rise time.
Y^2 = Yo^2 + 2g*h = 0.
28.6^2 + (-19.6)*h = 0,
h = 41.7 m. Above launching level.
ho - 0.5g*Tf^2 = 4.1.
41.7 - 4.9Tf^2 = 4.1,
Tf = 2.77 s. = Fall Time.
T = Tr+Tf = 2.92 + 2.77 = 5.69 s. = Time in air.
1 answer