At what angle should an object be thrown so that the maximum range is equal to three times the height reached by the projectile?

1 answer:

5 0

This link might help you: https://www.quora.com/At-what-angle-to-the-horizontal-should-a-body-be-projected-so-that-maximum-height-reached-is-equal-its-horizontal-range

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A web page designer creates an animation in which a dot on a computer screen has a position of r⃗ =[ 4.50 cm +( 2.90 cm/s2 )t2]i

VMariaS [17]

Answer:

V = (5.8cm/s)i, (4.7cm/s)j

Explanation:

Given :

r⃗ =[ 4.50 cm +( 2.90 cm/s2 )t2]i^+( 4.70 cm/s )tj^

To obtain the average velocity (V)

V = (r2 - r1) / (t2 - t1)

To obtain r1 and r2, substitute t1 = 0 and t2 = 2 respectively in the equation above

r1 = [ 4.50 cm +( 2.90 cm/s2 ) 0]i^+( 4.70 cm/s )0 j

r1 = 4.50 cm + 0 + 0 = (4.50cm)i + 0j

r2 = [ 4.50 cm +( 2.90 cm/s2 )2²]i^+( 4.70 cm/s )2 j

r2 = 4.50cm + (2.90 × 4)i + (4.70 × 2)j

r2 = (16.1cm)i + (9.4cm)j

V = [(16.1 - 4.50)i - (9.4 - 0)j] / 2 - 0

V = 11.6i / 2 ; 9.4j / 2

V = (5.8cm/s)i, (4.7cm/s)j

5 0

3 years ago

For an answer to be completed the units need to be specified why?

Setler79 [48]

Answer:

so you know the units in the problem and answer are the same, if they aren't you have to convert them to the same units.

3 0

3 years ago

A cheetah is walking at 1.0 m/s when it sees a zebra 25 m away. what acceleration would be required to reach 20.0 m/s in that di

inna [77]

Let
a = acceleration.
v = Speed
r = position
t = time.
To find the answer, you must write the equations of speed, and position known and clear the variables that are needed to find the acceleration.
The answer is a = 7.98m / s ^ 2.
I attach the solution.

8 0

3 years ago

The elevator accelerates upward (in the positive direction) from rest at a rate of 1.95 m/s2 for 2.15 s. Calculate the tension i

sammy [17]

The mass is missing. The mass of the elevator is 1650 kg.

Answer:

The tension in the cable is 19387.5 N.

Explanation:

Given:

Initial velocity of the elevator (u) = 0 m/s

Acceleration in the upward direction (a) = 1.95 m/s²

Time taken by the elevator (t) = 2.15 s

Mass of the elevator and persons (m) = 1650 kg

Let the tension in the cable wire be 'T' Newtons.

Now, there are 2 forces acting in the vertical direction. One is tension in the upward direction and the other the weight of the elevator in the downward direction.

As the elevator is accelerating upward, the net force acts in the upward direction.

So, net force on the elevator is given as:

$F_{net}=T-mg$