Answer:
27.3 m/s
Explanation:
We are given that
Distance travel by ball=x=60 m
[tex]\theta=26^{\circ}[/tex]
We have to find the initial speed([tex]v_0)[/tex] of the ball.
[tex]x=v_0cos\theta t[/tex]
Using the formula
[tex]60=v_0cos 26 t[/tex]
[tex]t=\frac{60}{v_ocos 26}=\frac{60}{v_0\times 0.899}=\frac{66.7}{v_0}[/tex]
The value of y at point of foot of the vertical distance
y=0
[tex]y=v_0sin\theta t-\frac{1}{2}gt^2[/tex]
Using [tex]g=9.8m/s^2[/tex]
Using the formula
[tex]0=v_0sin 26\times \frac{66.7}{v_0}-4.9\times (\frac{66.7}{v_0})^2[/tex]
[tex]4.9\times \frac{(66.7)^2}{v^2_0}=0.44\times 66.7[/tex]
[tex]v^2_0=\frac{4.9\times (66.7)^2}{0.44\times 66.7}[/tex]
[tex]v^2_0=742.8[/tex]
[tex]v_0=\sqrt{742.8}=27.3 m/s[/tex]
Hence, the initial speed of the ball=27.3 m/s
Answer:
27.3 m/s
Explanation:
Horizontal range, R = 60 m
angle of projection, θ = 26°
Let the velocity of projection is vo.
Use the formula of range of the projectile
[tex]R = \frac{u^{2}Sin2\theta} {g}[/tex]
[tex]60 = \frac{v_{0}^{2}Sin52}{9.8}[/tex]
vo = 27.3 m/s
Thus, the velocity of projection is 27.3 m/s.
