Question

A body is thrown with the velocity of 40m/s in direction making an angle of 30° with the horizontal. Calculate the time talen to reach the maximum height.

Answer 1

Range formula is :

Range = U²Sin 2Ф / g
U = initial velocity

Ф = angle of projectile

g = acceleration due to gravity

In our case Ф = 30°
U = 40 m/s
g = 10
Doing the substitution :
40² Sin (2 × 30) / 10 = 1600 Sin 60 / 10

1600 × 0.8660 = 1385.60

1385.60 / 10 = 138.560m

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Answer 2

$time \: taken = 2\frac{ {u} \sin( \theta ) }{g} \\ = \frac{2 \times 40 \times \sin(30) }{10} \\ = 4sec$
$height = \frac{ {(u \sin(\theta) )}^{2} }{2g} \\ = \frac{40 \times 40 \times {sin}^{2}(30) }{20} \\ 40m$
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