A projectile is fired v an initial rate of 75.2 m/s in ~ an edge of 34.5 degrees over the horizontal.Find:

a. Best height

b. Total horizontal distance


You should do your own homework problems. Gift told answers will certainly not assist you learn.

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Think about this. The tiger"s speed increases at a price of 9.81 m/s^2. For consistent acceleration, street is related to time through the equation d=tv_0+(t^2)a whereby v_0 is the initial speed IN THE DIRECTION that ACCELERATION, d is the street traveled, a is the rate of acceleration and also t is the time. Usage that to figure out exactly how much time that takes the tiger to reach the ground. Then multiply this lot of time by the front speed. This speak you how much the tiger has actually moved front upon landing.

For the sphere the very same formula applies. It works for the projectile too, but you need to decompose the velocity into its vertical and horizontal materials (hint: one will certainly be the sine the the edge it"s fired from times the speed and the various other will be the cosine of that angle time the speed). I"m walking to leave it as much as you to figure out which. This concern obviously has an ext to execute with wanting her homework answered because that you than with personal curiosity, due to the fact that of that is specificity, and also if you just didn"t get the principles it"s ok. The is not ok come submit her homework on the web in wishes that others will carry out it for you.


To uncover this answer, we have to multiply the tiger"s velocity through time in the air. Because gravity pulls at a continuous rate of 9.81m/s on earth, we use that as acceleration. Next, we find the time things takes to hit the ground.

distance = 1/2 acceleration time * time, therefore

7.5 = 1/2 9.81 t^2

And fix for t:

t^2 = 15/9.81

t = sqrt(15/9.81)

t = 1.2365

Now, time multiply by velocty

1.2365 * 4.5 = 5.5645

A ball is thrown vertically increase at 6.5 m/s. Exactly how high does that go?

So here, velocity is 6.5m/s and also acceleration is 9.81m/s

At the elevation of a balls increase path, it"s velocity have to equal 0. So, we must number out how long it takes because that the round to with a velocity that 0, then usage that to calculate height.

Divide the velocity by acceleration

6.5 / 9.81 = 0.6626 seconds

Now, if the ball deccelerates native 6.5 come 0 at a constant rate, the typical speed is 0.5 * 6.5, or using a formula:

v = 0.5 * V, so:

v = 0.5 * 6.5

v = 3.25 m/s

Now, to deal with for full distance, multiply mean velocity by time.

3.25 * 0.6626 = 2.1535 meter high

A projectile is fired through an initial speed of 75.2 m/s at an angle of 34.5 degrees over the horizontal.Find:

a. Maximum height

For this problem, we should use trigonometric attributes to calculation each velocity.

Vertical velocity: sin(34.5) 75.2 = 0.566 75.2 = 42.534

Horizontal velocity: cos(34.5) 75.2 = 0.824 75.2 = 61.695

So currently we usage the initial vertical velocity the 42.534 come calculate complete airtime.

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42.534 / 9.81 = 4.336 seconds


Average velocity is 0.5 * 42.534 = 21.267

21.267 * 4.336 = 92.214 meters high

b. Complete horizontal distance

We just calculated it to take it 4.336 secs to with maximum height, thus it additionally takes 4.336 seconds to reach the ground. 4.336 * 2 = 8.672 seconds total airtime

8.672 * 61.695 = 537.360 meters far


Here you carry out the math. The tiger will remain aloft precisely the exact same amount that time as it would certainly take a rock to drop come the ground once dropped from the offered height that 7.5 meters. Usage the formula because that gravational acceleration to calculate the time. Take her answer in seconds and multiply through the 4.5 meters per second of horizontal speed. Yes sir - so job-related it out.


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