The ball must be thrown at an angle of approximately 53.1∘ above the horizontal.
The trajectory of a projectile can be analyzed using principles of projectile motion. When a ball is thrown, it follows a curved path under the influence of gravity. To just clear a wall of height h, the vertical component of the initial velocity (v0sin(θ)) must be sufficient to overcome the height of the wall.
The vertical motion of the projectile is described by the equation h=v0sin(θ)t− 1/2gt^2, where h is the height of the wall, v0 is the initial velocity, θ is the angle of projection, t is the time of flight, and g is the acceleration due to gravity.
The time of flight (t) can be expressed as t= 2v0 sin(θ)/g . By substituting this expression for t into the vertical motion equation, we can solve for the angle (θ) that allows the ball to just clear the wall.
For a wall height of 5 meters and an initial velocity of 20 m/s, solving the equation yields an angle of approximately 53.1∘. This means that to clear the 5-meter-high wall, the ball should be thrown at an angle of approximately 53.1∘ above the horizontal, ensuring that the vertical component of the initial velocity is enough to reach and surpass the wall height.