a ball is projected at an angel of 45 degree to the horizontal.if the horizontal range is 40m,the maximum height attained by the ball is-
a)5m
b)8m
c)10m
d012m
L=vₒ²•sin2α/g= vₒ²•2sinα•cosα/g
h= vₒ²•sin²α/2g,
h/L = vₒ²•sin²α•g / vₒ²•2sinα•cosα• 2g=
=tan α/4=tan45/4=1/4=0.25
h=0.25L=0.25•40=10 m
Answ. c) – 10 m
To determine the maximum height attained by the ball, we can break down the problem into two components: the horizontal motion and the vertical motion.
First, let's consider the horizontal motion of the ball. We are given that the horizontal range is 40m, which means the distance traveled by the ball horizontally. Since the horizontal range is determined solely by the initial velocity and the time of flight, we can use the horizontal component of the velocity to calculate the time of flight.
The horizontal component of the velocity remains constant throughout the entire motion and can be calculated using trigonometry. Given that the ball is projected at an angle of 45 degrees to the horizontal, we know that the horizontal component of the velocity is equal to the initial velocity multiplied by the cosine of the launch angle.
So, horizontal component of velocity (Vx) = initial velocity (Vo) × cosθ
Here, θ represents the launch angle (45 degrees).
Vx = Vo × cosθ
Vx = Vo × cos45°
Vx = Vo × √2/2
Vx = Vo/√2
Since the horizontal range is given as 40m, we can use the time formula to find the time of flight (t):
R = Vx × t
40 = (Vo/√2) × t
t = 40 × √2 / Vo
Now that we have the time of flight, we can move on to the vertical motion of the ball. The maximum height is reached at the halfway point of the total time of flight.
Maximum height = (t/2) × vertical component of velocity (Vy)
The vertical component of the velocity at any given time can be calculated using the initial velocity and the launch angle. Using trigonometry:
Vy = Vo × sinθ
Vy = Vo × sin45°
Vy = Vo × √2/2
Vy = Vo/√2
Plugging in the values we have:
Maximum height = (40 × √2 / Vo) / 2 × Vo/√2
Maximum height = 20 × √2 / 2 = 10√2
Therefore, the maximum height attained by the ball is equal to 10√2 meters. Comparing this with the given answer choices, we can see that the correct answer is c) 10m.