Skip to content

#horizontal range

8 public questions tagged with this topic.

A projectile is launched at 36m/s at 37∘. What is its horizontal range? (Take g\=12m/s2,sin⁡74∘\=0.96)

Projectile motion splits into horizontal uniform and vertical accelerated motion per NCERT Chapter 4. Range R=u²sin2θ/g and max height H=u²sin²θ/2g. Using given u, θ, g, calculation yields 103.7 m. Hence option A satisfies projectile formulas.

Ref: NCERT Class 11 Physics > Chapter 4: Motion in a Plane > Topic: Relative Motion and Plane Motion

In projectile motion, what is the condition for achieving maximum horizontal range on level ground?

Projectile motion splits into horizontal uniform and vertical accelerated motion per NCERT Chapter 4. Range R=u²sin2θ/g and max height H=u²sin²θ/2g. Using given u, θ, g, calculation yields Launch angle of 45°. Hence option D satisfies projectile formulas.

Ref: NCERT Class 11 Physics > Chapter 4: Motion in a Plane > Topic: Two-Dimensional Motion and Vectors

A ball is thrown horizontally at 6m/s from a height of 4.9m. What is its horizontal range? (Take g\=9.8m/s2)

Time to fall: h=12gt2⇒4.9=12×9.8×t2⇒4.9=4.9t2⇒t2=1⇒t=1s. As per NCERT Chapter 3, equations v=u+at, s=ut+½at², v²=u²+2as describe uniformly accelerated motion. Applying correct signs and units gives 6 m as the result, so option D is correct.

Ref: NCERT Class 11 Physics > Chapter 3: Motion in a Straight Line > Topic: Motion Parameters and Calculations

A stone is thrown horizontally at 9m/s from a height of 78.4m. What is its horizontal range? (Take g\=9.8m/s2)

Time to fall: h=12gt2⇒78.4=12×9.8×t2⇒78.4=4.9t2⇒t2=16⇒t=4s. As per NCERT Chapter 3, equations v=u+at, s=ut+½at², v²=u²+2as describe uniformly accelerated motion. Applying correct signs and units gives 36 m as the result, so option D is correct.

Ref: NCERT Class 11 Physics > Chapter 3: Motion in a Straight Line > Topic: Straight Line Motion - Mixed Concepts

A ball is thrown horizontally at 15m/s from a height of 98m. What is its horizontal range? (Take g\=9.8m/s2)

Time to fall: h=12gt2⇒98=12×9.8×t2⇒98=4.9t2⇒t2=20⇒t=20≈4.47s. As per NCERT Chapter 3, equations v=u+at, s=ut+½at², v²=u²+2as describe uniformly accelerated motion. Applying correct signs and units gives 60 m as the result, so option A is correct.

Ref: NCERT Class 11 Physics > Chapter 3: Motion in a Straight Line > Topic: Straight Line Motion - Mixed Concepts

A ball is thrown horizontally at 5m/s from a height of 20m. What is its horizontal range? (Take g\=10m/s2)

Time to fall: h=12gt2⇒20=12×10×t2⇒20=5t2⇒t2=4⇒t=2s. As per NCERT Chapter 3, equations v=u+at, s=ut+½at², v²=u²+2as describe uniformly accelerated motion. Applying correct signs and units gives 10 m as the result, so option D is correct.

Ref: NCERT Class 11 Physics > Chapter 3: Motion in a Straight Line > Topic: Straight Line Motion - Mixed Concepts