Question

A projectile is fired with an initial speed of 36.4 m/s at an angle of 43.0 ∘ above the horizontal on a long flat firing range.

A) Determine the maximum height reached by the projectile.

B) Determine the total time in the air.

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

D)Determine the speed of the projectile 1.50 s after firing.

Answer 1

Part A.

Maximum Height in projectile motion is given by:

H_max = V0²*sin²A/(2*g)

Given that, V0 = 36.4 m/sec, A = Projectile angle = 43.0 deg above horizontal

g = 9.81 m/sec^2

So,

H_max = 36.4^2*(sin 43.0 deg)^2/(2*9.81)

H_max = 31.4 m/sec

Part B.

Total time of projectile motion is given by:

T = (2*V0*sin A)/g

Using given values:

T = (2*36.4*sin 43 deg)/9.81

T = 5.06 sec = total time in the air

Part C.

Range of projectile motion is given by:

R = V0_x*T

V0_x = Initial Horizontal velocity = V0*cos A

R = V0*T*cos A

R = 36.4*5.06*cos 43 deg

R = 134.7 m

Part D.

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

at t = 1.50 sec after firing

V1_x = V0_x

V1_x = V0*cos A = 36.4*cos 43 deg = 26.62 m/sec

Now In vertical direction there is always acceleration due to gravity downwards, So Using 1st kinematic equation:

V1_y = V0_y + a*t

a = -g = -9.81 m/sec^2 (-ve sign acceleration is in downward direction)

V0_y = V0*sin A = 36.4*sin 43 deg

t = 1.50 sec, So

V1_y = 36.4*sin 43 deg - 9.81*1.50

V1_y = 10.11 m/sec

Now speed of projectile at t = 1.50 sec will be:

|V1| = sqrt (V1_x^2 + V1_y^2)

|V1| = sqrt (26.62^2 + 10.11^2)

|V1| = 28.5 m/sec

Let me know if you've any query.