Question

A projectile is fired with an initial speed of 36.3 m/s at an angle of 41.0 degrees above the horizontal on a long flat firing range.

Determine the maximum height reached by the projectile.

Determine the total time in the air.

Determine the total horizontal distance covered (that is, the range).

Determine the speed of the projectile 2.00 s after firing.

Please provide readable steps to get a thumbs up. Thank you!

Answer 1

Given that Projectile is fired at V0 = 36.3 m/sec at 41.0 deg above the horizontal

V0x = Initial horizontal velocity = V0*cos = 36.3*cos 41.0 deg = 27.40 m/s

V0y = Initial vertical velocity = V0*sin = 36.3*sin 41.0 deg = 23.82 m/s

A.

Max height in projectile motion is given by:

H = V0y^2/(2g) = 23.82^2/(2*9.81)

H = 28.92 m

B.

time spent by projectile in air will be:

T = 2*V0y/g = 2*23.82/9.81

T = 4.86 sec

Part C.

Range in projectile motion is given by:

R = V0x*T

R = 27.40*4.86

R = 133.16 m

Part D.

Using 1st kinematic equation in vertical direction after 2.00 sec

Vfy = V0y + ay*T

Vfy = 23.82 + (-9.81)*2.00

Vfy = 4.2 m/s (+ve sign means vertical velocity will be in upward direction)

Since there is no acceleration in horizontal direction, So horizontal velocity will remain constant

Vfx = V0x = 27.40 m/s

final velocity of projectile will be:

Vf = Vfx i + Vfy j

Vf = (27.40 i + 4.2 j) m/s

Magnitude of the velocity will be:

|Vf| = sqrt (Vfx^2 + Vfy^2)

|Vf| = sqrt (27.40^2 + 4.2^2)

|Vf| = 27.72 m/s = Speed of projectile after 2.00 s

Let me know if you've any query.