All categories

3 years ago

Jumps at 8 m/s at a 30 degree angle above the horizontal. how long was she in the air?

1 answer:

4 0

Vo = 8 m/s. g = 10 m/s^2
alpha = 30 degree
ask: x?

x = vo^2 sin ^2 alpha / g
x = 8 ^2 x 2 sin 30 cos 30 / 10
x =( 64 x2x 1/2x1/2√3 )/ 10
x= 3.2 √3 metres

√3 = 1.73

You might be interested in

What does an atomic nucleus give off a particle?

____ [38]

The correct answer is answer choice B.

5 0

3 years ago

Give THREE examples of objects that have potential energy

Tasya [4]

A ball kept on 3rd floor of a building.

A pendulum bob kept at 3m height

A stone thrown vertically upward.

A pressed spring.

A squashed spunge ball.

4 0

3 years ago

A short-wave radio antenna is supported by two guy wires, 155 ft and 175 ft long. Each wire is attached to the top of the antenn

Vladimir79 [104]

Answer:

163.8 ft

Explanation:

In triangle ABD

$AB$ = 155 ft

$Cos63 = \frac{BD}{AB} = \frac{BD}{155}\\BD = 155 Cos63 \\BD = 70.4 ft$

$Sin63 = \frac{AD}{AB} = \frac{AD}{155} \\AD = 166 Sin63\\AD = 148 ft$

Using Pythagorean theorem in triangle ADC

$AC^{2} = AD^{2} + DC^{2} \\175^{2} = 148^{2} + DC^{2} \\DC = 93.4 ft$

$d$ = distance between the anchor points

distance between the anchor points is given as

$d = BD + CD = 70.4 + 93.4\\d = 163.8 ft$

5 0

3 years ago

Read 2 more answers

A car is traveling at a velocity of 22 m/s when the driver puts on the brakes

Brums [2.3K]

The car’s velocity at the end of this distance is 18.17 m/s.

Given the following data:

Initial velocity, U = 22 m/s

Deceleration, d = 1.4 $m/s^2$

Distance, S = 110 meters

To find the car’s velocity at the end of this distance, we would use the third equation of motion;

Mathematically, the third equation of motion is calculated by using the formula;

$V^2 = U^2 + 2dS$

Substituting the values into the formula, we have;

$V^2 = 22 + 2(1.4)(110)\\\\V^2 = 22 + 308\\\\V^2 = 330\\\\V^2 = \sqrt{330}$

Final velocity, V = 18.17 m/s

Therefore, the car’s velocity at the end of this distance is 18.17 m/s.