A projectile is projected with speed 49 ms-1 at an angle of 30° with horizontal. The time taken by the projectile to reach the highest point is

Question

A projectile is projected with speed 49 ms-1 at an angle of 30° with horizontal. The time taken by the projectile to reach the highest point is (A) 2 s (B) 2.5 s (C) 3.5 s (D) 4.5 s

Tardigrade Admin · Accepted Answer

Here, Velocity of projection, u=49 m s-1 Angle of projection, θ=30° Time taken by the projectile to reach the highest point is t=(u sin θ/g) =((49 m s-1)(sin 30°)/9.8 m s-2) =2.5 s

Q. A projectile is projected with speed $49 m s^{- 1}$ at an angle of $3 0^{\circ}$ with horizontal. The time taken by the projectile to reach the highest point is

$2 s$

$2.5 s$

$3.5 s$

$4.5 s$

Here,
Velocity of projection, $u = 49 m s^{- 1}$
Angle of projection, $θ = 3 0^{\circ}$
Time taken by the projectile to reach the highest point is
$t = \frac{u s in θ}{g}$
$= \frac{( 49 m s ^{- 1} ) ( s in 3 0 ^{\circ} )}{9.8 m s ^{- 2}}$
$= 2.5 s$

Q. A projectile is projected with speed 49ms−1 at an angle of 30∘ with horizontal. The time taken by the projectile to reach the highest point is

2s

2.5s

3.5s

4.5s