Question

A ball is projected at an angle 45 to the horizontal. If the horizontal range is 20,the maximum height to which the ball rises​

Answer 1

Answer:

Equation of projectile  is 

y=ntanθ(1−Rn)

Total distance the ball covers before falling =10+40=50 m

∴Range =50m

Now, y=10tan450(1−5010)

=10×1×5040=2×4=8m

Height of wall =8 m

Answer 2

Answer:

The maximum height reached by the ball is 5 meters.

Step-by-step explanation:

Range = (Vi^2 * sin2θ) / g where g = 9.8 m/s^2 and θ = 45 degrees

Maximum Height = (Vi^2 * sin^2 θ) / 2g

Using the range formula

Range = (Vi^2 * sin2θ) / g

20 m = (Vi^2 * sin 2 * 45) / g

20 m = (Vi^2 ) * sin 90) / g

20 m = Vi^2 * 1) / g

20 m = Vi^2 / g

Solving for Vi^2

Vi^2 = 20 m * g

Solving for the maximum height, maxH of the ball

maxH = (Vi^2 * sin^2 45) / 2g

maxH = (Vi^2 * 0.5) / 2g

maxH = (20 m * g * 0.5) / 2 g

maxH = (10 m * g) / 2g

maxH = 5 m