Answer:
[tex]v_{i}=16.4m/s[/tex]
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
[tex]v_{x}=vicos\alpha \\v_{ti}=v_{i}sin\alpha[/tex]
The horizontal distance is given by
[tex]x=v_{x}t[/tex]
put value of x and vx we get
[tex]x=v_{x}t\\10=v_{i}cos\alpha t\\10=v_{i}cos25^{o} t\\11=v_{i}t\\[/tex]
11=vit...........(1)
The vertical displacement is given by:
[tex]y=v_{yi}t+1/2gt^{2}[/tex]
substitute given values
so
[tex]2.44=(v_{i}t)sin(25)+(1/2)(-9.8)t^{2}\\[/tex]......(2)
From eq(1) and (1)
we get
[tex]t=0.67s[/tex]
So the initial velocity we get from eq(1)
[tex]11=v_{i}t\\11=v_{i}*0.67\\v_{i}=11/0.67\\v_{i}=16.4m/s[/tex]
Answer:
16.4 m/s
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
