Question

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

Answer 1

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.

Answer 2

steve are you mean take the quadratic of this -4.9t^2+28.63t-4.1

Answer 3

your syntax is strange, but yes, the quadratic formula is the best way to find the desired time.

Answer 4

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.