Question

A soccer goal is 2.44 m high. A player kicks the ball at a distance 10 m from the goal at an angle of 25°. What is the initial speed of the soccer ball?

Answer 1

16.4 m/s

Step-by-step explanation:

With the assumption that the ball passes through the frame of goal

y = 2.44 m

x = 10 m

θ = 25°

u = initial velocity of the ball = ?

But, we can divide the initial velocity into horizontal and vertical components and use equations of motion to analyse this.

uₓ = u cos 25° = 0.9063 u

uᵧ = u sin 25° = 0.4226 u

The horizontal distance covered during the flight of a projectile is given by

R = uₓt (no acceleration part, since there's no acceleration in the horizontal direction)

10 = 0.9063 u × t

u = (11/t)

The vertical component of the ball's position is given by

y = uᵧt + (1/2)gt²

y = 2.44 m, uᵧ = 0.4226 u = 0.4226 (11/t) = (4.65/t), g = - 9.8 m/s²

2.44 = 4.65 - 4.9 t²

t² = 0.451

t = 0.672 s

u = (11/t) = 11/0.672 = 16.4 m/s

Answer 2

`v_(i)=16.4m/s`

Step-by-step explanation:

Given data

y=2.44 m

x=10 m

α=25°

To find vi

Solution

The components of initial velocity vi is given as

`v_(x)=vicos\alpha \\v_(ti)=v_(i)sin\alpha`

The horizontal distance is given by

`x=v_(x)t`

put value of x and vx we get

`x=v_(x)t\\10=v_(i)cos\alpha t\\10=v_(i)cos25^(o) t\\11=v_(i)t\\`

11=vit...........(1)

The vertical displacement is given by:

`y=v_(yi)t+1/2gt^(2)`

substitute given values

so

`2.44=(v_(i)t)sin(25)+(1/2)(-9.8)t^(2)\\`......(2)

From eq(1) and (1)

we get

`t=0.67s`

So the initial velocity we get from eq(1)

`11=v_(i)t\\11=v_(i)*0.67\\v_(i)=11/0.67\\v_(i)=16.4m/s`