Question

An object is launched at a velocity of 20 m/s in a direction making an angle of 25 degrees upward with the horizontal.

a) What is the maximum height reached by the object?

b) What is the total flight time (between launch and touching the ground) of the object?

c) What is the horizontal range (maximum x above ground) of the object?

d) What is the magnitude of the horizontal component of the velocity of the object just before it hits the ground?

e) What is the magnitude of the vertical component of the velocity of the object just before it hits the ground?

f) What is the magnitude of the total velocity of the object just before it hits the ground?

g) What is the angle with which the object hits the ground?

Answer 1

Given that

Initial velocity of projectile = v0 at angle with horizontal

Horizontal component of velocity, v0x = v0*cos = 20*cos 25 deg = 18.13 m/s

Vertical component of velocity, v0y = v0*sin = 20*sin 25 deg = 8.45 m/s

Since there is no acceleration in horizontal direction, So horizontal velocity of projectile remains constant.

acceleration in vertical direction = -g = -9.81 m/s^2

Part A.

At max height vertical velocity component will be zero, So

Using 3rd kinematic equation

V1y^2 = V0y^2 + 2*ay*Hmax

Hmax = (V1y^2 - V0y^2)/(2*ay)

Hmax = (0^2 - 8.45^2)/(2*(-9.81))

Hmax = 3.64 m

Part B.

flight time in projectile motion is given by:

T = 2*V0y/g

T = 2*8.45/9.81

T = 1.72 sec

Part C.

Range in projectile motion is given by:

R = V0x*T

R = 18.13*1.72

R = 31.18 m

Part D.

Since there is no acceleration in horizontal direction, So horizontal velocity of projectile remains constant.

Vfx = V0x

Vfx = 18.13 m/s

Part E.

Using 1st kinematic equation in vertical direction:

Vfy = V0y + ay*T

Vfy = 8.45 + (-9.81)*1.72

Vfy = -8.45 m/s

Part F.

final velocity will be:

Vf = Vfx i + Vfy j

Vf = 18.13 i - 8.45 j

Magnitude of velocity will be:

|Vf| = sqrt ((18.13)^2 + (-8.45)^2)

|Vf| = 20.0 m/s = final speed

Part G.

Direction = arctan (Vfy/Vfx)

Direction = arctan (-8.45/18.13)

Direction = -25.0 deg = 25.0 deg below the horizontal

Let me know if you've any query.