A ball is thrown at an angle $53^{\circ}$ with a velocity of 50 $m s^{- 1}$ . Calculate the difference between the velocities at the two instants when the body is 20 m above the ground

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C
$V_{x} = u c o s 53^{\circ} = 30 m s^{- 1} v_{y}^{2} = {(u s i n θ)}^{2} - 2 g h \Rightarrow v_{y} = \pm 20 \sqrt{3} m s^{- 1} ∴ {\to v}_{1} = 30^i + 20 \sqrt{3}^j a n d {\to v}_{2} = 30^i - 20 \sqrt{3}^j ∴ Δ v = {\to v}_{1} - {\to v}_{2} = 40 \sqrt{3}^j m s^{- 1}$

Suggest Corrections

Similar questions

The velocity of point A on the rod is 2 $m s^{- 1}$ (leftwards) at the instant shown in fig. The velocity of the point B on the rod at this instant is

A body of mass $5 k g$ is moving along the x-axis with a velocity $2 m s^{- 1}$ . Another body of mass $10 k g$ is moving along the y-axis with a velocity $\sqrt{3} m s^{- 1}$ . They collide at the origin and stick together. The final velocity of the combined mass is

A projectile is fired at $30^{\circ}$ to the horizontal. The vertical component of its velocity is $80 m s^{- 1}$ . Its time of flight is T. What will be the velocity of the projectile at $t = \frac{T}{2}$

Q. A man holds an umbrella at

30^{\circ}

with the vertical to keep himself dry. He, then, runs at a speed of

10 {ms}^{- 1}

and finds the rain drops to be hitting vertically. Speed of the rain drops w.r.t. the running man and w.r.t. earth are

Q. A body is projected with a velocity of

30 m/s

at an angle of

60^{\circ}

to the horizontal. Find the magnitude of velocity when the angle between the velocity vector and acceleration vector is

120^{\circ}

Join BYJU'S Learning Program

A ball is thrown at an angle 53∘ with a velocity of 50 ms−1. Calculate the difference between the velocities at the two instants when the body is 20 m above the ground

The correct option is C Vx=u cos 53∘=30ms−1v2y=(u sinθ)2−2gh⇒vy=±20√3 ms−1∴→v1=30^i+20√3^jand →v2=30^i−20√3^j∴Δv=→v1−→v2=40√3^j ms−1

A ball is thrown at an angle $53^{\circ}$ with a velocity of 50 $m s^{- 1}$ . Calculate the difference between the velocities at the two instants when the body is 20 m above the ground

The correct option is C
$V_{x} = u c o s 53^{\circ} = 30 m s^{- 1} v_{y}^{2} = {(u s i n θ)}^{2} - 2 g h \Rightarrow v_{y} = \pm 20 \sqrt{3} m s^{- 1} ∴ {\to v}_{1} = 30^i + 20 \sqrt{3}^j a n d {\to v}_{2} = 30^i - 20 \sqrt{3}^j ∴ Δ v = {\to v}_{1} - {\to v}_{2} = 40 \sqrt{3}^j m s^{- 1}$