Question

(ii) The ball has a time of flight of 3.5s before landing The green is a vertical
distance of 7.5 m above the point at which the ball was hit, as shown. The green
is about seventy metres away from where the ball is hít.
flight of ball
green
ball hit at this point
17.5m
7 Om
Deduce whether, if air resistance is ignored, the ball could land on the green
after a flight time of 3,5s.

Answer 1

If air resistance is ignored, the ball could land on the green after a flight time of 3.5s.

Vertical velocity of the ball

The vertical velocity of the ball is calculated as follows;

$h = v_yt - \frac{1}{2} gt^2\\\\7.5 = (3.5)v_y - \frac{1}{2}(9.8)(3.5)^2\\\\ 7.5 = 3.5v_y - 60.025\\\\3.5v_y = 67.525\\\\v_y = 19.29 \ m/s$

Horizontal velocity of the ball

X = Vxt

Vx = (X)/t

Vx = (70)/3.5

Vx = 20 m/s

Resultant initial velocity

v = √(Vy² + Vx²)

v = √(19.29² + 20²)

v = 27.79 m/s

Angle of projection

tanθ = (Vy)/(Vx)

tanθ = (19.29)/(20)

tanθ = 0.9645

θ = 43.96⁰

check if time of flight will be equal to 3.5 s or more

The time of flight is calculated as follows;

$T = \frac{2u \times sin(\theta)}{g} \\\\T = \frac{2 \times 27.79 \times sin(43.96)}{9.8} \\\\T = 3.9 \ s$

Check if the range of ball will 70 m or more

X = VxT

X = 20 x 3.9

X = 78 m

The ball will land on green which is at 70 m after 3.5s.

Thus, if air resistance is ignored, the ball could land on the green after a flight time of 3.5s.

Learn more about time of flight here: https://brainly.com/question/4441382

#SPJ1