Physics-projectiles calculate maximum height ymax? please help?

Question

need help please

A projectile was fired 2936 metres across level ground at an initial angle of 52.00 degrees above the horizontal. What was the maximum height of this projectile?

Give your answer in metres (m) and round it to the nearest metre.

Maximum Height (m) =

Answer 1

First you have to draw your picture!  That is key.  You end up getting a right triangle with the hypotenuse being 2936 m/s, and the angle measures are 52º (the one from the cannon) 90º, and 38º (the angle in the air).  From here, use trigonometry to solve for the missing sides of the triangle.  To find the vertical side is sin(52) = x/2936 (sin equals opposite over hyp.).  From there, you get the vertical speed to be exactly 2936sin(52).  Now take your Physics formula vf^2 = vi^2 + 2ad to solve for maximum height:

0^2 = (2936sin(52))^2 + 2(-9.8)d

Acceleration due to gravity is negative because it is acting against the bullet.

Solve algebraically to get your answer:

distance = 273,099.1317 m

rounded to the nearest meter, your answer is 273,099 m.

Hope this helps ;).

Answer 2

Initial velocity (u) = 2936 m/sec

Θ = 52.00 degrees

g = 9.8 m/sec²

Let the maximum height be x

We have

x = u² . Sin²Θ / 2.g

=> x = (2936)² . Sin² 52 / 2 * 9.8

=> x = 8620096 * (0.788)² / 19.6

=> x = 8620096 * 0.6209 / 19.6

=> x = 273072.326 m

or x = 273072 meters

Answer 3

use trigonometry - so tangent(angle) = opposite/adjacent therefore divide distance by 2 then times it by tangent 52

Answer 4

For this speed the difference in g at different altitudes cannot be neglected.

Use conservation of energy as your main equation, with

KE=(1/2)mv^2

and

PE=-GMm/r

The initial r is the radius of the Earth, the final r is what is being solved for (sort of, we'll get there).  The initial speed is what's given, the final speed is

v_f=v_i cos(theta)

Set up the equation, solve for r_f (this is a little ugly, but not bad).

Once you have r_f, subtract r_i (the radius of the Earth) to get the altitude.