Question

A ball of 600 grams is kicked at an angle of 35° with the ground with an initial velocity V0.
a) What is the initial velocity V0 of the ball if its kinetic energy is 22 Joules when its height is maximum?
b) What is the maximum height reached by the ball

Answer 1

Answer:

hola mate

here is ur answer

a) 

a) When the height of the ball is maximum, the vertical component of its velocity is zero; hence the kinetic energy is due to its horizontal component Vx = V0 cos (θ). 

a) When the height of the ball is maximum, the vertical component of its velocity is zero; hence the kinetic energy is due to its horizontal component Vx = V0 cos (θ). 22 = (1/2) m (Vx)2

a) When the height of the ball is maximum, the vertical component of its velocity is zero; hence the kinetic energy is due to its horizontal component Vx = V0 cos (θ). 22 = (1/2) m (Vx)222 = (1/2) 0.6 (V0 cos (35°))2

a) When the height of the ball is maximum, the vertical component of its velocity is zero; hence the kinetic energy is due to its horizontal component Vx = V0 cos (θ). 22 = (1/2) m (Vx)222 = (1/2) 0.6 (V0 cos (35°))2V0 = (1 / cos (35°)) √(44/0.6) = 10.4 m/s 

a) When the height of the ball is maximum, the vertical component of its velocity is zero; hence the kinetic energy is due to its horizontal component Vx = V0 cos (θ). 22 = (1/2) m (Vx)222 = (1/2) 0.6 (V0 cos (35°))2V0 = (1 / cos (35°)) √(44/0.6) = 10.4 m/s b) 

a) When the height of the ball is maximum, the vertical component of its velocity is zero; hence the kinetic energy is due to its horizontal component Vx = V0 cos (θ). 22 = (1/2) m (Vx)222 = (1/2) 0.6 (V0 cos (35°))2V0 = (1 / cos (35°)) √(44/0.6) = 10.4 m/s b) Initial kinetic energy (just after the ball is kicked) 

a) When the height of the ball is maximum, the vertical component of its velocity is zero; hence the kinetic energy is due to its horizontal component Vx = V0 cos (θ). 22 = (1/2) m (Vx)222 = (1/2) 0.6 (V0 cos (35°))2V0 = (1 / cos (35°)) √(44/0.6) = 10.4 m/s b) Initial kinetic energy (just after the ball is kicked) Ei = (1/2) m V02 = (1/2) 0.6 (10.4)2 = 32.4 J 

a) When the height of the ball is maximum, the vertical component of its velocity is zero; hence the kinetic energy is due to its horizontal component Vx = V0 cos (θ). 22 = (1/2) m (Vx)222 = (1/2) 0.6 (V0 cos (35°))2V0 = (1 / cos (35°)) √(44/0.6) = 10.4 m/s b) Initial kinetic energy (just after the ball is kicked) Ei = (1/2) m V02 = (1/2) 0.6 (10.4)2 = 32.4 J The difference between initial kinetic energy and kinetic energy when the ball is at maximum height H is equal to gain in potential energy 

a) When the height of the ball is maximum, the vertical component of its velocity is zero; hence the kinetic energy is due to its horizontal component Vx = V0 cos (θ). 22 = (1/2) m (Vx)222 = (1/2) 0.6 (V0 cos (35°))2V0 = (1 / cos (35°)) √(44/0.6) = 10.4 m/s b) Initial kinetic energy (just after the ball is kicked) Ei = (1/2) m V02 = (1/2) 0.6 (10.4)2 = 32.4 J The difference between initial kinetic energy and kinetic energy when the ball is at maximum height H is equal to gain in potential energy 32.4 - 22 = m g H 

a) When the height of the ball is maximum, the vertical component of its velocity is zero; hence the kinetic energy is due to its horizontal component Vx = V0 cos (θ). 22 = (1/2) m (Vx)222 = (1/2) 0.6 (V0 cos (35°))2V0 = (1 / cos (35°)) √(44/0.6) = 10.4 m/s b) Initial kinetic energy (just after the ball is kicked) Ei = (1/2) m V02 = (1/2) 0.6 (10.4)2 = 32.4 J The difference between initial kinetic energy and kinetic energy when the ball is at maximum height H is equal to gain in potential energy 32.4 - 22 = m g H H = 10.4 / (0.6 * 9.8) = 1.8 m

tq❤

Answer 2

Answer:

a)The initial velocity of the ball is 11.31m/s

b)The maximum height reached by the ball is 2.14m

Explanation:

Given that a ball of mass 600 grams is kicked at an angle of 35° with the ground with an initial velocity V0.

a)We are required to find the initial velocity of the ball if the kinetic energy of ball at maximum height is 22J.The kinetic energy of the ball is

$K.E=\frac{1}{2}mv^{2}$

The mass of ball is

$m = 0.6 \: kg$

Only vertical component of velocity exists at maximum height and it is

$V_{0}sin35°$

Substituting these values in kinetic energy relation,we get

$22 = \frac{1}{2} \times 0.6 \times (V_{0}sin35°) {}^{2} \\ = > 44 = 0.6 \times V_{0} {}^{2} \times0 .573 \\ = > V_{0} {}^{2} = \frac{44}{0.6 \times 0.573} = 127.98 \\ = > V_{0} = 11.31ms {}^{ - 1}$

Therefore,the initial velocity of the ball is 11.31m/s

b)The formula for finding maximum height reached by the ball is

$H_{max}=\frac{V_{0}^{2}sin^{2}\theta}{2g}$

Substituting the known values in above relation,we get

$H_{max}=\frac{11.31^{2}sin^{2}35}{2 \times 9.81} \\ = \frac{11.31 {}^{2} \times {0.573}^{2} }{19.62} = 2.14 \: m$

Therefore,the maximum height reached by the ball is 2.14m

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