A ball is thrown with an initial speed of 20 m/s at an angle of 60° to the ground. If air resistance is negligible, what is the ball’s speed at the instant it reaches its maximum height from the ground?.



Answer :

The speed of the ball when it reaches its maximum height from the ground is 5.02 m/s.

The initial speed of the ball = 20 m/s

Angle to the at which the ball is thrown = 60°

Gravity = 9.81 m/s²

Resistance of the air = 0.

The maximum height reached by the ball above the ground is,

[tex]H = \frac{u ^{2 }sin ^{2}θ} {2g}[/tex]

[tex]H = \frac{20 ^{2 }sin ^{2}60°} {2 \times 9.81}[/tex]

[tex]H = \frac{ 20 \times 20 \times \sqrt{3} }{2 \times 9.81 \times 2}[/tex]

[tex] = 17.6 \: m[/tex]

Thus, the maximum height reached by the ball above the ground is 17.6 m.

The flight time of the ball is,

[tex]t = \frac{2usinθ}{g} [/tex]

[tex]t = \frac{2 \times 20 \times sin60°}{9.81} [/tex]

[tex]t = \frac{2 \times 20 \times \sqrt{3} }{9.81 \times 2} [/tex]

[tex]t = 3.5 \: s[/tex]

Thus, the flight time of the ball is 3.5 s.

The speed of the ball when it reaches it reaches the maximum height from the ground is,

[tex]Speed = \frac{ Distance }{Time}[/tex]

[tex]Speed = \frac{17.6}{3.5} [/tex]

[tex]Speed = 5.02 \: m/s[/tex]

Therefore, the speed of the ball when it reaches its maximum height from the ground is 5.02 m/s.

To know more about projectile motion, refer to the below link:

https://brainly.in/question/14665257

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