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Anika [276]
3 years ago
9

A projectile is fired with an initial velocity of 450 feet per second at an angle of 70° with the horizontal.

Physics
1 answer:
pav-90 [236]3 years ago
5 0
<h2>After 26.28 seconds projectile returns 26.28 seconds.</h2>

Explanation:

Initial velocity = 450 ft/s = 137.16 m/s

Angle, θ = 70°

Consider the vertical motion of projectile,

When the projectile return to the ground we have

           Displacement, s = 0 m

           Acceleration, a = -9.81 m/s²

            Initial velocity, u = 137.16 x sin70 = 128.89 m/s

Substituting in s = ut + 0.5 at²

                s = ut + 0.5 at²

                0 = 128.89 x t + 0.5 x (-9.81) x t²

                t² - 26.28 t = 0

                t ( t- 26.28) = 0

               t = 0 s or t = 26.28 s

After 26.28 seconds projectile returns 26.28 seconds.

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3 0
3 years ago
A ball is thrown vertically upwards with a velocity of 30m/s. Determine the maximum height reached
padilas [110]
The maximum height reached is 45.92 m

6 0
2 years ago
The question is in the picture
Sedbober [7]

Answer:

e) 120m/s

Explanation:

When the ball reaches its highest point, its velocity becomes zero, meaning

v_0-gt = 0.

where v_0 is the initial velocity.

Solving for t we get

t = \dfrac{v_0}{g}

which is the time it takes the ball to reach the highest point.

Now, after the ball has reached its highest point, it turns around and falls downwards. After time t_0 since it had reached the highest point, the ball has traveled downwards and the velocity v_f it has gained is

v_f = gt_0,

and we are told that this is twice the initial velocity v_0; therefore,

v_f = 2v_0  = gt_0

which gives

t_0 = \dfrac{2v_0}{g}.

Thus, the total time taken to reach velocity 2v_0 is

t_{tot} = t+t_0 = \dfrac{v_0}{g}+\dfrac{2v_0}{g}

t_{tot} = \dfrac{3v_0}{g}.

This t_{tot}, we are told, is 36 seconds; therefore,

36= \dfrac{3v_0}{g},

and solving for v_0 we get:

v_0 = \dfrac{36g}{3}

v_0 = \dfrac{36s(10m/s^2)}{3}

\boxed{v_0 = 120m/s}

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7 0
3 years ago
You are on a train that is traveling at 3.0 m/s along a level straight track. Very near and parallel to the track is a wall that
loris [4]

Questions Diagram is attached below

Answer:

T=2.08s

Explanation:

From the question we are told that:

Speed of Train V=3.0m.s

Angle \theta=12\textdegree

Height of window h_w=0.90m

Width of window w_w=2.0m

The Horizontal distance between B and A from Trigonometric Laws is mathematically given by

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Therefore

Distance from A-A

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Therefore

Time Required to travel trough d is mathematically given as

 T=\frac{d_a}{v}

 T=\frac{6.23}{3}

 T=2.08s

5 0
2 years ago
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