Question

javelin is thrown at an angle of 30 degree with an initial speed of 25 m/s. (given g = 10m/s^2) What is the maximum Height

Answer 1

Answer:

Given : u=20 m/s θ=30

o

g=10 m/s

2

We get sinθ=sin30

o

=

2

1

Maximum height reached H=

2g

u

2

sin

2

θ

=

20

20×20×(

2

1

)

2

=5m

Answer 2

Given :

javelin is thrown at an angle of 30°

Initial speed = 25m/s

To find :

The maximum height of Javelin

Solution :

The maximum height is given as,

$\boxed { \gray{\bf H = \dfrac{ {u}^{2} { sin}^{2} \theta }{2g} }}$

where,

• H denotes maximum height
• u denotes initial velocity
• g denotes gravity

According to question,

θ = 30°

u = 25m/s

by substituting all the given values in the equation,

${: \implies{\sf H = \dfrac{ {u}^{2} { sin}^{2} \theta }{2g} }}$

${ :\implies{\sf H = \dfrac{ {(25)}^{2} \times { \bigg( \dfrac{1}{2} \bigg)}^{2} }{2 \times 10} }}$

${: \implies{\sf H = \dfrac{ 625 \times \dfrac{1}{4} }{20} }}$

${: \implies{\sf H = \dfrac{ 156.25 }{20} }}$

${: \implies \underline { \boxed{ \purple{\sf H = 7.81 \: m }}}}$

thus, the maximum height of the javelin is 7.81m