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Pani-rosa [81]
3 years ago
9

A ball is kicked at an angle of 35° with the ground.a) What should be the initial velocity of the ball so that it hits a target

that is 30 meters away at a height of 1.8 meters?b) What is the time for the ball to reach the target?
Physics
1 answer:
stiks02 [169]3 years ago
7 0

Answer:

a.18.5 m/s

b.1.98 s

Explanation:

We are given that

\theta=35^{\circ}

a.Let v_0 be the initial velocity of the ball.

Distance,x=30 m

Height,h=1.8 m

v_x=v_0cos\theta=v_0cos35

v_y=v_0sin\theta=v_0sin35

x=v_0cos\theta\times t=v_0cos35\times t

t=\frac{30}{v_0cos35}

h=v_yt-\frac{1}{2}gt^2

Substitute the values

1.8=v_0sin35\frac{30}{v_0cos35}-\frac{1}{2}(9.8)(\frac{30}{v_0cso35})^2

1.8=30tan35-\frac{6574.6}{v^2_0}

\frac{6574.6}{v^2_0}=21-1.8=19.2

v^2_0=\frac{6574.6}{19.2}

v_0=\sqrt{\frac{6574.6}{19.2}}=18.5 m/s

Initial velocity of the ball=18.5 m/s

b.Substitute the value then we get

t=\frac{30}{18.5cos35}

t=1.98 s

Hence, the time for the ball to reach the target=1.98 s

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