Projectile Motion Problem: Finding Maximum Height and Horizontal Range

Given:
Initial velocity (v) = 98 m/s
Angle of projection (θ) = 60 degrees
Acceleration due to gravity (g) = 9.8 m/s^2

To Find:
Maximum height attained by the body
Horizontal range covered by the body

Solution:

Step 1: Break the initial velocity into horizontal and vertical components:
The initial velocity (v) can be resolved into horizontal and vertical components using trigonometry.

Horizontal component: vx = v cos θ = 98 cos 60 = 49 m/s
Vertical component: vy = v sin θ = 98 sin 60 = 84.85 m/s

Step 2: Determine the time taken for the body to reach maximum height:
The time taken for the body to reach maximum height can be calculated using the vertical component of the initial velocity (vy) and the acceleration due to gravity (g).

Using the formula: vy = u + gt
where u is the initial velocity, t is the time taken, and g is the acceleration due to gravity.

At maximum height, the vertical component of velocity becomes zero (v = 0).

0 = vy - gt
t = vy/g = 84.85/9.8 = 8.65 seconds

Step 3: Calculate the maximum height reached by the body:
The maximum height reached by the body can be calculated using the formula:

h = uy*t - ½ * g * t^2
where uy is the initial vertical velocity and t is the time taken to reach maximum height.

h = 84.85*8.65 - ½*9.8*(8.65)^2 = 359.8 meters

Therefore, the maximum height attained by the body is 359.8 meters.

Step 4: Calculate the horizontal range:
The horizontal range covered by the body can be calculated using the horizontal component of the initial velocity (vx) and the time taken for the body to reach maximum height (t).

Using the formula: R = vx * t

R = 49 * 8.65 = 423.85 meters

Therefore, the horizontal range covered by the body is 423.85 meters.

Conclusion:
The maximum height attained by the body is 359.8 meters and the horizontal range covered by the body is 423.85 meters.

U = 98m/s
thetha (∆) = 60

Max. height = (u^2 sin^2∆)2g

So, Max height = (98 ×98 × (sin60)^2 )/ 2 × 9.8
= (98 ×98 × 3/4)/ 2×9.8

On solving
Max height = 367.5


Horizontal range = (u^2 sin 2∆)/g
= (98×98×√3/2)/9.8

on solving. .
R=848.7