# A projectile is fired from ground level with an initial speed of 300 m/sec and an angle of elevation of 30 degrees. Use that the acceleration due to g

By On Saturday, June 8th, 2019 Categories : Question & Answer

A projectile is fired from ground level with an initial speed of 300 m/sec and an angle of elevation of 30 degrees. Use that the acceleration due to g. Are You mam and sir has this kind of inquiry?, If do then plz found the good tips below this line:

Using projectile formulas in physics we can solve the initial velocity by the formula R=Vx(t)
Where Vx is the horizontal component of the velocity and `t` is the time elapsed after the projectile reaches the distance R, where Vx is also equals to 2Vo(cosineØ/g
and g=9.8m/sec/sec (the acceleration due to gravity)
and
t=2Vo(sineØ/g
where Vo is the initial velocity and Ø is the angle of projectile which is 45degrees…
Substituting in the formulas and applying algebraic simplification and trigonometic identities will result to the simple equation of
R=Vo(squared) x sin2Ø /g

Hence
81.1m=Vo(squared) x sin90/(9.8m/sec/sec)
Therefore the initial velocity Vo can be computed approx. as
Vo=28.2m/sec

The time `t` would then be
t=2x(28.2)sin45deg./g
=4.07 seconds

And if the projectile angle would be change other than 45 degrees the projectile would reach at a shorter distance because the sine of 90degrees is the only value that reaches the maximum value of unity or 1 any other angle has the value of less than 1..hope I answered all your questions,tnx!

Dan_Olila

Okay let us solve your problem. Lets start with given values, building erected from the ground creates a right angle, that means we have a right triangle that can be imagined. Here is a link of a picture i made for you simple so you can image it, it is easier to solve the height of the building.

In the image you will see the height of the building is missing as well as the supposedly hypotenuse.

Since you have one side given which is the shadow of the building which is 21meters and having all the angles you can solve the height by using the sine, cosine, and tangent.

If you are not sure and is new to this solution you can have this to remember SOHCAHTOA.
Legend:
S=Sine C=Cosine T=Tangent
O=Opposite side to angle
H=Hypothenuse

we will rely on givens to solve problems
SOH=sine angle = opposite/hypothenuse
CAH=Cosine angle given = adjacent side/hypotenuse
TOA=Tangent given angle = opposite side/adjacent side

For this problem we will use TOA since there is an angle given and an adjacent side given and we are looking for the opposite side.

Tangent 57 degrees = opposite side/21 meters
1.5=opposite/21meters
opposite side = 1.5x21meters
opposite site = 31.5 meters
the height of the building is 31.5 meters

Note: i have rounded up the values to solve the problem.

mikem20

One of the equation of motions for a projectile motion is expressed as Vy = Voy – gt, where Vy is the horizontal velocity; Voy is the initial horizontal velocity, g = 9.8m/s^2; and, t is the time. For an initial velocity of 73.9m/s fired at 65.6 above the horizontal, the initial horizontal velocity would be Voy = 73.9m/s * sin(65.6). Note that at the maximum height, the horizontal velocity of the shell would be zero. Thus, Vy = 0. From the given values, you can solve the time, t, when the shell would be at its maximum. Using the solved t value, you can obtain the maximum height from

y = (Voy * t) – (g*t^2)

pinm21

Thank you for visiting us here.

The maximum height the projectile can go is: 470.204 m

Below is the solution for reference and guidance.

Vy = 96 m/s ; Vx = 940 m/s ; g = 9.8 m/s^2

V^2= Vx^2 + Vy^2
V = 944.89 m/s

V = V1 – gt
0 = 96 – 9.8 t
t = 9.796 s

h = (1/2) gt^2
= 0.5 * 9.8 * (9.796)^2
= 470.204 m

I hope this helps you.

MARKtheSPARK47

clownz

Yes,when an object moves through a fluid(gas or liquid),it always experiences a friction force which is called drag force in opposite direction to the motion of object.this force varies with the speed of object upto certain limit.

hamza ali

Solution:
Use inverse trigonometric function:
θ = tan-1 (Opposite Leg/