Question

A ball is thrown at an angle of 45° from horizontal its velociy is 20m/s.(a) Find the time taken to stike the ground.(b)Find the maximum height it reaches.(c)How far away from the point its throws?[Take g=10m/s^2]​

Answer 1

${\bold{\underline{\underline{Answer:}}}}$

${\bold{\therefore Time=2\sqrt{2}\:sec}}$

${\bold{\therefore h_{max}=10\:m}}$

${\bold{\therefore Range=40\:m}}$

${\bold{\underline{\underline{Step-by-step\:explantion:}}}}$

• In the given question information given about a ball is thrown at an angle of 45° from horizontal its velociy is 20m/s.

• We have to (a)Find the time taken to stike the ground.(b)Find the maximum height it reaches.(c)How far away from the point its throws?

$\underline \bold{Given : } \\ \implies Initial \: velocity(u) = 20 \: m/s \\ \\ \implies Angle \: of \: projection( \theta) = 45 \degree \\ \\ \underline \bold{to \: find: } \\ \implies Time \:of \: flight(T) = ? \\ \\ \implies Maximum \: height(h_{max} )= ? \\ \\ \implies Range(R) = ?$

• According to given question :

$\bold{For \: time \: of \: flight : } \\ \\ \bold{In \: y \: direction : } \\ \implies s= u_{y}t + \frac{1}{2} {a _{y} t}^{2} \\ \\ \implies 0 = u \: sin \theta \times t - \frac{1}{2} {gt}^{2} \\ \\ \implies T= \frac{2u \: sin \theta}{g} \\ \\ \implies T= \frac{2 \times 20 \times \frac{1}{ \sqrt{2} } }{10} \\ \\ \implies t = \frac{\cancel{20} \sqrt{2} }{\cancel{10}} \\ \\ \bold{\implies T= 2\sqrt{2} \: sec} \\ \\ \:\:\:\:\bold{For \: h_{max} : } \\ \\ \bold{in \: y \: direction: } \\ \implies {v}^{2} _{y} - {u}^{2} _{y} = 2{a} _{y}.dy \\ \\ \implies {0}^{2} - {(u \: sin \theta)}^{2} = 2( - g) h_{max} \\ \\ \implies h_{max} = \frac{ {u}^{2} sin^{2} \theta }{2g} \\ \\ \implies h_{max} = \frac{\cancel{20} \times \cancel{20 }\times \frac{1}{\cancel2} }{\cancel{2 }\times \cancel{10}} \\ \\ \bold{\implies h_{max} = 10 \: m} \\ \\ \: \: \: \: \: \: \: \: \: \bold{For \: Range : } \\ \implies Range= \frac{ {u}^{2} \: sin 2 \theta }{g} \\ \\ \implies Range = \frac{20 \times \cancel{20} \times 1}{\cancel{10}} \\ \\ \bold{\implies Range = 40 \: m}$

Answer 2

Answer:

[a] Time taken=$2\sqrt{2}\:sec$

[b] Maximum height= 10 m

[c] Range= 40 m

Step-by-step explanation :

$\bold{given } \\ \to velocity \: of \: ball = 20 \: ms \\ \to angle = 45 \degree \\ \\ \bold{Using \: formula \: of \: time \: of \: flight : } \\ \to t = \frac{2usin \theta}{g} \\ \\ \to t = \frac{2 \times 20 \times \frac{1}{ \sqrt{2} } }{10} \\ \\ \to t = \frac{20 \sqrt{2} }{10} \\ \\ \to t = 2 \sqrt{2 } \: sec \\ \\ \bold{Using \: formula \: of \: h_{max} }\\ \to h_{max} = \frac{ {u}^{2} {sin}^{2} \theta }{2g} \\ \to h_{max} = \frac{20 \times 20 \times \frac{1}{2} }{2 \times 10} \\ \\ \to h_{max} = 10 \: m \\ \\ \bold{Using \: formula \: of \: \: range} \\ \to r = \frac{ {u}^{2}sin \: 2 \: \theta }{g} \\ \\ \to r = \frac{20 \times 20 \times 1}{10} \\ \\ \to r = 40 \: m$