IDNLearn.com provides a reliable platform for finding accurate and timely answers. Join our interactive community and access reliable, detailed answers from experienced professionals across a variety of topics.
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
#SPJ4