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defon
4 years ago
8

A projectile is launched at a 30° angle relative to the ground. The projectile has an initial velocity of 15 m/s and travels thr

ough the air for 2 seconds.
What is the horizontal displacement (rounded to the nearest hundredth) in m?
Physics
1 answer:
Salsk061 [2.6K]4 years ago
7 0

Answer:

The horizontal displacement of the projectile is 25.98 m

Explanation:

Given;

angle of projection, θ = 30°

initial velocity of the projectile, v = 15 m/s

time of flight, t = 2 seconds

The horizontal displacement or range is given by;

R = vₓt

where;

vₓ is the horizontal component of the velocity

t is the time of flight

R = (15cos30)(2)

R = 25.981 m

R = 25.98 m (to the nearest hundredth)

Therefore, the horizontal displacement of the projectile is 25.98 m

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Answer:

Explanation:

The tendency of the automobile running on a circular path at high speed to turn towards left or right around one of its wheels , is due to torque by centripetal force acting at its centre of mass about that wheel .

Suppose the automobile tends to turn in clockwise direction about its wheel on the outer edge . A rotational angular momentum is created . To counter this effect , we can take the help of a rotating wheel or flywheel . We shall have to keep its rotation in anticlockwise direction so that it can create rotational angular momentum in direction opposite to that created by centrifugal force on fast moving automobile. Each of them will nullify the effect of the other . In this way , rotating flywheel will save the automobile from turning upside down .

4 0
4 years ago
An astronaut on the moon throws a baseball upward. The astronaut is 6​ ft, 6 in.​ tall, and the initial velocity of the ball is
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Answer: 89.803 ft

Explanation:

The complete question is written below:

An astronaut on the moon throws a baseball upward. The astronaut is 6ft, 6 in. tall and the initial velocity of the ball is 30 ft per second. The height s of the ball in feet is given by the equation,s=-2.7t^2+30t+6.5, where t is the number of seconds after the ball is thrown.

The ball will never reach a height of 100ft. How can this be determined algebraically?

We have the following equation that expresses the height s as a function of time:

s=-2.7t^{2}+30t+6.5 (1)

Now, if we wan to find the maximum height the baseball reaches and prove it is less than 100 ft, we firstly have to find the time t_{total} the whole parabolic movement lasts and then find t_{smax}=\frac{t_{total}}{2} which is the time it takes the baseball to reach its maximum height.

So, if we want to calculate t_{total}, this is fulfilled when s=0, when the baseball hits the ground:

0=-2.7t^{2}+30t+6.5 (2)

This is a quadratic equation of the form 0=at^{2}+bt+c, and we have to use the quadratic formula if we want to find  t_{total}:

t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}  (3)

Where a=-2.7, b=30, c=6.5

Substituting the known values and choosing the positive result of the equation:

t_{total}=11.323 s  (4)

Then we can calculate t_{smax}:

t_{smax}=\frac{t_{total}}{2} (5)

t_{smax}=\frac{11.323 s}{2}

t_{smax}=5.661 s  (6)

Substituting (6) in (1):

s=-2.7(5.661 s)^{2}+30(5.661 s)+6.5 (7)

s=89.803 ft (8) This is the maximum height the baseball reaches, as we can see it is less than 100 ft

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Answer:

Explanation:

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