Question

A body is projected by making an angle 30 degree with the horizontal with a velocity of 39.2 m/s ,find

a)Time of flight(T)

b)Range (R)

c)Maximum height (H)

Answer 1

Step-by-step explanation:

Maximum height,

H = u² sin²β/2g = 30²× sin²30°/(2×10) = 11.25 m

Time of flight ,

T = 2u sinβ /g = 2×30 ×sin30°/10 =3 sec

Horizontal range,

R= u²sin2β/g = 30² × sin60°/10 = 45√3 m

Answer 2

(a). The time of flight is 2 sec.

(b). The range is 135.79 m.

(c). The maximum height is 19.6 m.

Step-by-step explanation:

Given that,

Angle = 30°

Horizontal velocity = 39.2 m/s

(a). We need to calculate the time of flight

Using formula of time of flight

$t=\dfrac{u\sin\theta}{g}$

Put the value into the formula

$t=\dfrac{39.2\times\sin30}{9.8}$

$t=2\ sec$

(b). We need to calculate the range

Using formula of range

$R=\dfrac{u^2\sin2\theta}{g}$

Put the value into the formula

$R=\dfrac{39.2^2\sin60}{9.8}$

$R=135.79\ m$

(c). We need to calculate the maximum height

Using formula of maximum height

$h_{max}=\dfrac{u^2\sin^2\theta}{2g}$

Put the value into the formula

$h_{max}=\dfrac{39.2^2\sin^2(30)}{2\times9.8}$

$h_{max}=19.6\ m$

Hence, (a). The time of flight is 2 sec.

(b). The range is 135.79 m.

(c). The maximum height is 19.6 m.

Learn more :

Topic : projectile motion

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