Respuesta :
Answer:
(a) Maximum height = 53.88 meters
(b) Range of the ball = 924.36 meters
Explanation:
The ball has been launched at a speed = 44.4 meters per second
Angle of the ball with the horizontal = 25°
Horizontal component of the speed of the ball = 44.4cos25° = 40.24 meters per second
Vertical component = 44.4sin25° = 18.76 meters per second
We know vertical component of the speed decides the height of the ball so by the law of motion,
v² = u² - 2gh
where v = velocity at the maximum height = 0
u = initial velocity = 18.76 meter per second
g = gravitational force = 9.8 meter per second²
Now we plug in the values in the given equation
0 = (18.76)² - 2(9.8)(h)
19.6h = 352.10
h = [tex]\frac{352.10}{19.6}[/tex]
h = 17.96 meters
By another equation,
[tex]v=ut-\frac{1}{2}gt^{2}[/tex]
Now we plug in the values again
[tex]0=(18.76)t-\frac{1}{2}(9.8)t^{2}[/tex]
18.76t = 4.9t²
18.76 = 4.9t
t = [tex]\frac{18.76}{4.9}=3.83[/tex]seconds
Since time t is the time to cover half of the range.
Therefore, time taken by the ball to cover the complete range = 2×3.83 = 7.66 seconds
So the range of the ball = Horizontal component of the velocity × time
= 40.24 × 7.66
= 308.12 meters
This we have calculated all for our planet.
Now we take other planet.
(a) Since the golfer drives the ball 3 times as far as he would have on earth then maximum height achieved by the ball = 17.96 × 3 = 53.88 meters
(b) Range of the ball = 3×308.12 = 924.36 meters
