A cricketer can throw a ball to maximum horizontal distance of 160 m. Calculate the maximum vertical hight to which Hi can tharow the Ball given g=10 Ms-2
Answers
Answer:
80m
Explanation:
So we know that R or range = 160m
We know apply the formula for calculating range of projectile motion, where "u" is initial velocity, "g" is acceleration due to gravity and "alpha" is the angle at which the ball is thrown.
For maximum horizontal distance, the cricketer must throw the ball at an angle of 45°.
\frac{ {u}^{2} \sin(2 \alpha ) }{g} = 160
g
u
2
sin(2α)
=160
We now substitute the value of alpha in our equation.
\begin{gathered}\frac{ {u}^{2} \sin(90) }{g} = 160 \\ \frac{ {u}^{2}}{g} = 160\end{gathered}
g
u
2
sin(90)
=160
g
u
2
=160
We know rearrange the following equation of motion in the given way with appropriate signs, where "v" if the final velocity, "u" and "g" are the same as before and "s" is the distance travelled.
\begin{gathered}{v}^{2} - {u}^{2} = 2( - g)s \\ - {u}^{2} = - 2gs \\ \frac{ {u}^{2} }{2g} = s\end{gathered}
v
2
−u
2
=2(−g)s
−u
2
=−2gs
2g
u
2
=s
And we substitute the value from the previous equation to get the value of "s", which is now the maximum vertical distance which we had to find.
\begin{gathered}\frac{160}{2} = s \\ s = 80\end{gathered}
2
160
=s
s=80
Hence the answer is 80m.
Answer:
40m
Explanation:
R = 2u^2/g
160= 2u^2/10
u^2 = 800m/sec
h=u^2/2g
= 800/20
h =40m.