Question

A soccer player kicks a ball at an angle of 36◦ from the direct horizontal with an initial speed of 15.5 m/s. Assuming that the ball moves in a two-dimensional plane, find its maximum height and its velocity when it strikes the ground.

Answer 1

Launching speed = 15.5 m/sec at 36 deg above the horizontal

Max height in projectile motion is given by:

H_max = (V0^2*sin^2 )/(2*g)

Using above values:

H_max = (15.5^2*(sin 36 deg)^2)/(2*9.81)

H_max = 4.23 m

Part B.

Time taken by projectile to land will be given by:

T = (2*V0*sin )/g

T = (2*15.5*sin 36 deg)/9.81

T = 1.86 sec

Now since there is no acceleration in horizontal direction, So horizontal velocity remains constant,

Vfx = V0x

Vfx = V0*cos = 15.5*cos 36 deg = 12.54 m/s

In vertical direction using 1st kinematic equation:

Vfy = V0y + a*T

Vfy = V0*sin - g*T

Vfy = 15.5*sin 36 deg - 9.81*1.86 = -9.13 m/s

So final velocity when it strikes on the ground will be:

Vf = Vfx i + Vfy j

Vf = 12.54 i - 9.13 j

Magnitude of final velocity = |Vf| = sqrt (12.54^2 + (-9.13)^2) = 15.5 m/s

Direction = arctan (-9.13/12.54) = -36.0 deg

final velocity = 15.5 m/s at 36 deg below the horizontal

Let me know if you've any query.