Question

A man throws ball at maximum horizontal distance 80 m calculate maximum height reached

Answer 1

Here is your answer buddy ↙️↙️☺️☺️

We know that R = u2 * Sin 2θ / g.

We know that R = u2 * Sin 2θ / g.As, the range is maximum, so, θ = 450

We know that R = u2 * Sin 2θ / g.As, the range is maximum, so, θ = 450So, 80 = u2 /g.

We know that R = u2 * Sin 2θ / g.As, the range is maximum, so, θ = 450So, 80 = u2 /g.Now, H = height reached by ball = u2 Sin2θ / 2g.

We know that R = u2 * Sin 2θ / g.As, the range is maximum, so, θ = 450So, 80 = u2 /g.Now, H = height reached by ball = u2 Sin2θ / 2g.Putting value of θ = 450 , u2 / g = 80, we get

We know that R = u2 * Sin 2θ / g.As, the range is maximum, so, θ = 450So, 80 = u2 /g.Now, H = height reached by ball = u2 Sin2θ / 2g.Putting value of θ = 450 , u2 / g = 80, we getH = 80/4 = 20m

hope this helps you ☺️☺️❤️❤️✌️✌️

Answer 2

the maximum height it covers is

h/2= 80/2=40 m