A ball is projected from the ground at angle 60 from the ground, if the ball just clear a wall 5m from point of projection and falls on the ground 15m away from it on other side, then height of the wall is nearly equal to?

(1) 7.9m (2) 9.8m (3) 5.8m (4) 6.5m

To solve this problem, we need to analyze the projectile motion of the ball. We know that the ball is projected at an angle of 60 degrees from the ground, clears a wall that is 5 meters away, and lands 15 meters away from the point of projection. Our goal is to find the height of the wall.

Breaking Down the Problem

First, let's establish some key points:

• The angle of projection, θ = 60 degrees.
• The horizontal distance to the wall, x = 5 meters.
• The total horizontal distance traveled, R = 15 meters.

Understanding Projectile Motion

In projectile motion, the horizontal and vertical motions can be analyzed separately. The horizontal distance (range) and the height can be calculated using the initial velocity (u) and the angle of projection.

Calculating the Time of Flight to the Wall

The horizontal component of the initial velocity can be expressed as:

u_x = u * cos(θ)

The vertical component is:

u_y = u * sin(θ)

Since we know the horizontal distance to the wall (5 m), we can find the time (t) it takes to reach the wall:

x = u_x * t

Rearranging gives:

t = x / u_x = 5 / (u * cos(60))

Since cos(60) = 0.5, this simplifies to:

t = 10 / u

Finding the Height of the Wall

Next, we need to calculate the height (h) of the ball when it reaches the wall. The vertical motion can be described by the equation:

h = u_y * t - (1/2) * g * t²

Substituting for u_y and t:

h = (u * sin(60)) * (10 / u) - (1/2) * g * (10 / u)²

Since sin(60) = √3/2, we have:

h = 10 * (√3/2) - (1/2) * g * (100 / u²)

Using the Total Range to Find Initial Velocity

The total range (R) can also be expressed as:

R = (u_x * T)

Where T is the total time of flight. The total time can be derived from the vertical motion equations, but we can also use the range formula for projectile motion:

R = (u² * sin(2θ)) / g

Substituting θ = 60 degrees gives sin(120) = √3/2:

15 = (u² * √3) / g

Rearranging gives:

u² = (15 * g) / √3

Substituting Back to Find Height

Now we can substitute u² back into our height equation:

h = 10 * (√3/2) - (1/2) * g * (100 * √3 / (15 * g))

After simplifying, we find:

h = 5√3 - (10√3 / 3) = (15√3 - 10√3) / 3 = 5√3 / 3

Calculating this gives approximately 8.66 m. However, we need to consider the options provided.

Final Calculation and Answer

After evaluating the options, the closest height of the wall that the ball clears is:

• (1) 7.9m
• (2) 9.8m
• (3) 5.8m
• (4) 6.5m

Given our calculations, the height of the wall is nearly equal to 7.9m.