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2 years ago

14

If the ball shown in the figure lands in 1.0 s, about what height was it thrown from?

2 answers:

Scrat [10]2 years ago

6 0

Answer:

5m

Explanation:

Rus_ich [418]2 years ago

4 0

Answer:

Should be 5 m

Explanation:

hope it helps

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A basketball star covers 2.70 m horizontally in a jump to dunk the ball (see figure). His motion through space can be modeled pr

Likurg_2 [28]

Answer:

Part a)

$T = 0.81 s$

Part b)

$v_x = 3.33 m/s$

Part c)

$v_y = 3.91 m/s$

Part d)

$\theta = 49.55 degree$

Part e)

$T = 1.11 s$

Explanation:

Part a)

initial vertical position = 1.02 m

maximum height = 1.80 m

$\Delta y = 1.80 - 1.02$

$\Delta y = 0.78 m$

$v_f^2 - v_y^2 = 2a \Delta y$

$0 - v_y^2 = 2(-9.81)(0.78)$

$v_y = 3.91 m/s$

time taken by it to reach this height

$v_y = v_i + at$

$0 = 3.91 - 9.81 t_1$

$t_1 = 0.39 s$

Now when it again touch the ground then its speed is given as

$v_f^2 - v_y^2 = 2a \Delta y$

$v_f^2 - 0 = 2(9.81)(1.80 - 0.95)$

$v_y = 4.08 m/s$

time taken by it to reach this height

$4.08 = v_i + at$

$4.08 = 0 + 9.81 t_2$

$t_2 = 0.42 s$

$T = t_1 + t_2$

$T = 0.81 s$

Part b)

Horizontal velocity

$v_x = \frac{x}{t}$

$v_x = \frac{2.70}{0.81}$

$v_x = 3.33 m/s$

Part c)

vertical velocity is the intial y direction velocity

$v_y = 3.91 m/s$

Part d)

Take off angle is given as

$tan\theta = \frac{3.91}{3.33}$

$\theta = 49.55 degree$

Part e)

initial vertical position = 1.20 m

maximum height = 2.50 m

$\Delta y = 2.50 - 1.20$

$\Delta y = 1.30 m$

$v_f^2 - v_y^2 = 2a \Delta y$

$0 - v_y^2 = 2(-9.81)(1.30)$

$v_y = 5.05 m/s$

time taken by it to reach this height

$v_y = v_i + at$

$0 = 5.05 - 9.81 t_1$

$t_1 = 0.51 s$

Now when it again touch the ground then its speed is given as

$v_f^2 - v_y^2 = 2a \Delta y$

$v_f^2 - 0 = 2(9.81)(2.50 - 0.72)$

$v_y = 5.9 m/s$

time taken by it to reach this height

$5.9 = v_i + at$

$5.9 = 0 + 9.81 t_2$

$t_2 = 0.60 s$

$T = t_1 + t_2$

$T = 1.11 s$

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