A ball is projected with initial velocity 50m\s at an angle of elevation of 37°from the top of a cliff 55m high. calculate the total time the ball is in the air and the maximum horizontal distance covered...please I have solved this question but didn't get the answer ..I need ur assistance..
To calculate the total time the ball is in the air and the maximum horizontal distance covered, we can break down the motion of the ball into horizontal and vertical components.
1. Vertical motion:
Let's start by calculating the time taken for the ball to reach its maximum height. We can use the formula for vertical motion:
Formula: v = u + at
Here, u is the initial vertical velocity, v is the final vertical velocity (which will be 0 at the maximum height), a is the acceleration due to gravity (-9.8 m/s^2), and t is the time.
Given:
u (initial vertical velocity) = 50 m/s (since the ball is projected upward)
a (acceleration due to gravity) = -9.8 m/s^2 (which is negative as it acts downwards)
v = u + at
0 = 50 + (-9.8)t
Solving this equation gives:
50 = 9.8t
t = 50 / 9.8
t ≈ 5.1 seconds (rounded to one decimal place)
So it takes approximately 5.1 seconds for the ball to reach its maximum height.
Now, to find the total time of flight, we know that the time for the ascent is the same as the time for the descent. Thus, the total time in the air is twice the time taken to reach the maximum height.
Total time in the air = 2 * 5.1 seconds = 10.2 seconds
2. Horizontal motion:
To calculate the maximum horizontal distance covered by the ball, we can use the formula:
Formula: horizontal distance (d) = horizontal velocity (u) * time (t)
The initial horizontal velocity (u) can be determined using trigonometry. Since the ball is projected at an angle of 37°, we can find u using:
u = initial velocity * cos(angle of elevation)
Given:
initial velocity = 50 m/s
angle of elevation = 37°
u = 50 * cos(37°)
u ≈ 40.2 m/s (rounded to one decimal place)
Now, we can calculate the maximum horizontal distance (d) using:
d = u * t
d = 40.2 * 10.2
d ≈ 410.4 meters (rounded to one decimal place)
Therefore, the total time the ball is in the air is approximately 10.2 seconds, and the maximum horizontal distance covered is approximately 410.4 meters.
Sure, I would be happy to assist you.
To solve this problem, we can break it down into two components: vertical motion and horizontal motion.
Vertical motion involves calculating the time the ball takes to reach its maximum height (where its vertical velocity becomes zero), as well as the time it takes to return to the ground.
1. Calculating time to maximum height:
The vertical velocity component (Vy) of the initial velocity (v₀) can be found using trigonometry:
Vy = v₀ * sin(θ)
Vy = 50 * sin(37°)
Next, we can use the equation of motion to find the time (t₁) taken to reach the maximum height:
t₁ = Vy / g
where g is the acceleration due to gravity (~9.8 m/s²).
2. Calculating total time in the air:
Since the ball spends an equal amount of time on the ascent and descent, the total time in the air (t_total) is given by:
t_total = 2 * t₁
3. Calculating maximum horizontal distance:
The horizontal velocity component (Vx) of the initial velocity (v₀) can be found using trigonometry:
Vx = v₀ * cos(θ)
Vx = 50 * cos(37°)
To find the maximum horizontal distance, we need to determine the time taken for the ball to reach the ground (t₂). Since the initial vertical position is 55 m, the equation for vertical displacement is:
h = v₀ * sin(θ) * t - (1/2) * g * t²
where h is the vertical displacement.
Solving this equation for t, we get two solutions since the ball reaches the ground both during the ascent and descent. We can discard the negative solution since t cannot be negative:
t₂ = (-Vy - √(Vy² - 2g(-55))) / (-g)
Finally, we can calculate the maximum horizontal distance (D) covered by the formula:
D = Vx * (t_total or t₂)
Plugging in the known values, you should be able to find the correct answers.
the initial vertical velocity is
... 50 m/s * sin(37º)
the (vertical) freefall equation is
. h = 1/2 g t^2 + [50 sin(37º)] t + 55
plug in zero for h (ball on ground) and solve for t (use quadratic formula)
the horizontal velocity is
... 50 m/s * cos(37º)
... multiply by t to find distance
remember...g is negative (downward)