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Colt1911 [192]
3 years ago
13

In 1993, Wayne Brian threw a spear at a record distance of 201.24 m. (This is not an official sports record because a special de

vice was used to “elongate” Brian’s hand.) Suppose Brian threw the spear at a 35.0° angle with respect to the horizontal. What was the initial speed of the spear? 2. Find the maximum height and time of flight of the spear in problem #1.
I really don't know how to do any of this please help me :(
Physics
1 answer:
Tcecarenko [31]3 years ago
6 0

Answer:

V₀ = 45.81 m/s

H = 70.45 m

T = 5.36 s

Explanation:

The motion of the spare is projectile motion. Therefore, we will first use the formula of range of projectile:

R = V₀² Sin 2θ/g

where,

R = Range of Projectile = 201.24 m

V₀ = Initial Speed = ?

θ = Launch Angle = 35°

g = 9.8 m/s²

Therefore,

201.24 m = V₀²[Sin 2(35°)]/9.8 m/s²

V₀ = √[(201.24 m)/(0.095 m/s²)

<u>V₀ = 45.81 m/s</u>

Now, for maximum height:

H = V₀² Sin² θ/g

H = (45.81 m/s)² Sin² 35°/9.8 m/s²

<u>H = 70.45 m</u>

For the total time of flight:

T = 2 V₀ Sin θ/g

T = 2(45.81 m/s) Sin 35°/9.8 m/s²

<u>T = 5.36 s</u>

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Two concentric, coplanar, circular loops of wire of different diameter carry currents in the same direction. Describe the nature
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Answer:

Both will be attractive in nature.

Explanation:

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Why do we see the sun before it actually rises in the morning?
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3 years ago
If a ball that is 10 meters above the ground is thrown horizontally at 5.51 meters per second. a. how long will it take for the
GalinKa [24]

Answer:

a. t = 1.43 s

b. d = 7.88 m

Explanation:

a. The time of flight can be found using the following equation:

y_{f} = y_{0} + v_{0_{y}}t - \frac{1}{2}gt^{2}

Where:

y_{f}: is the final height = -10 m

y_{0}: is the initial height = 0

v_{0_{y}}: is the initial speed in the vertical direction = 0

g: is the acceleration due to gravity = 9.81 m/s²

By solving the above equation for "t" we have:

t = \sqrt{\frac{2y_{f}}{g}} = \sqrt{\frac{2*10 m}{9.81 m/s^{2}}} = 1.43 s

Hence, the ball will hit the ground in 1.43 s.

b. The distance in the horizontal direction can be found as follows:

x_{f} = x_{0} + v_{0}t + \frac{1}{2}at^{2}

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x₀: is the initial position in the horizontal direction = 0

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x_{f} = 5.51 m/s*1.43 s = 7.88 m

Therefore, the ball will travel 7.88 m before it hits the ground.

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