Question

A projectile is fired from ground level with an initial speed of 550 m/sec and an angle of elevation of 30 degrees. Use that the acceleration due to gravity is 9.8 m/sec2
What is the range of the projectile? ___ meters
What is the max height of the projectile? __ meters
what is the speed at which the projectile hits the ground? ___m/sec

Answer 1

Answer:

x (max) = 26760 m

y (max)  = 3859  meters

V = 549.5  m/sec

Step-by-step explanation:

Equations to describe the projectile shot movement are:

a(x)  =  0     V(x)  =  V(₀) *cos α                  x  =  V(₀) *cos α  * t

a(y)  = -g     V(y)  =  V(₀) * sin α  - g*t         y  =   V(₀) * sin α *t  - (1/2)*g*t²

a ) What is the range of the projectile.   α  =  30°

then  sin 30° = 1/2     cos  30° = √3 /2   and     tan 30° = 1/√3

x  maximum occurs when in the equation of trajectory  we make  y  = 0

Then

y  =  x*tan α  -  g*x / 2*V(₀)²*cos²  α

x*tan α  =  g*x /  2* V(₀)²*cos²  α

By subtitution

1/√3    =   9.8* x(max)  / 2* (550)²*0.75

(1/√3) * 453750 / 9.8  = x (max)

x (max) = 453750 / 16.95   meters

x (max) = 26760 m

The maximum height is when V(y) = 0

We compute t in that condition

V(y)  =  0  =  V(₀) * sin α - g*t

t  =   V(₀) * sin α / g       ⇒   t  =  550* (1/2) / 9.8

t  =  28.06 sec

Then  h (max)  =   y(max)  =  V(₀) sin α * t  - 1/2 g* t²

y (max)  =  550* (1/2)*28.06 - (1/2)* 9.8 * (28.06)²

y (max)  =  7717  -  3858  

y (max)  = 3859  meters

What is the speed when the projectile hits the ground

V  =  V(x)  + V (y)    and   t  =  2* 28.06         t  =  56.12 sec

mod V  =√ V(x)²  + V(y)²

V(x)  =  V(₀) cos α    =  550 * √3/2

V(x)  = 475.5  m/sec       V(x)²   =  226338 m²/sec²

V(y) =  550*1/2   -  9.8* 56.12     ⇒  V(y) = 275  -  549.98

V(y) =  - 274.98          V(y) ²   =  

V =  √ 226338 + 75614     ⇒  V = 549.5  m/sec