Question

A cricket ball is thrown at a speed of 28m/s in a direction 30degree above the horizontal. Calculate i)aximum height II) time of flight III) horizontal range

Answer 1

maximum height is..
$u {}^{2} sin {}^{2} \alpha \div 2g$
28 ×28×sin^2 30/2×9.8
=>28×28/2 ×4×10
=>9.8

Time of flight is..
$2usin \alpha \div g$
2×28×sin30/10
=>2.8

Horizontal range is..
u^2sin2theta/g
=>28×28×sin2×60/10
=>39.2√3

Answer 2

Answer:A/Q

Given-initial velocity u=28m/s

SinΦ=30°

Therefore,

(a) The maximum height-

H=u^2*sin^2Φ/2g

  =28*28*(1/2)^2/2g

  =28*28*1/4/19.6

  =28*7/19.6

  =196/19.6

  =10m

(b) Time taken-

T=2u*sin30°/g

 =2*28*sin30°/g [since,sin30=1/2]

 =2*14/9.8 [2*28*1/2=2*14]

 =28/9.8

 =2.85

T=2.9[approximately]

Now, (c) Maximum range-

R=u^2*sin2Φ/g

 =28*28*sin60°/g

 =784*0.866/g [since, sin60°=0.866]

 =678.944/9.8

 =69.28m