A golfer hits a ball and gives it an initial velocity of 40 m/s, at an angle of 30 degrees above the horizontal. A) how long does the ball stay in the air? (Answer is 4.08 sec)

B) how far horizontally (the range) does the ball travel before hitting the ground? (Answer 141 m)

I need help with getting the work. If y’all could explain, that would be so helpful and thank you.

(a) The ball will stay in the air for 4.08 s.

(b) The ball travels 141 m horizontally before hitting the ground.

This is a projectile motion problem, to solve the problems above, we need to use the formula of (a) Time of flight and (b) Range.

(a)

⇒ Formula

  T = 2Usin∅/g Equation 1

⇒Where:

T = Time of flightU = initial velocity∅ = Angle above the horizontalg = acceleration due to gravity

From the question,

⇒Given:

U = 40 m/s∅ = 30°

⇒Assuming:

g = 9.8 m/s²

⇒Substitute these values into equation 1

t = 40sin30/9.8t = 2(40)(0.5)/9.8t = 40/9.8t = 4.08 s

Also,

(b) For Range

⇒ Formula

R = U²sin2∅/g Equation 2

⇒Substitute these values above into equation 2

R = 40²sin(2×30)/9.8R = 1600×0.866/9.8R = 141.38 mR ≈ 141 m

Hence, (a) The ball will stay in the air for 4.08 s (b) The ball travels 141 m horizontally before hitting the ground

Learn more about projectile motion here: link

Ibelieve the correct answer would be c.

Does this your question?