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MCQEasyJEE 2026Projectile Motion

JEE Physics 2026 Question with Solution

A boy throws a ball into air at 45∘45^\circ from the horizontal to land it on a roof of a building of height HH. If the ball attains maximum height in 2 s2 \, \text{s} and lands on the building in 3 s3 \, \text{s} after launch, then the value of HH is _____ m\text{m}.

(Given: g=10 m s−2g = 10 \, \text{m s}^{-2})

  • A

    2525

  • B

    1010

  • C

    1515

  • D

    2020

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: Time to reach maximum height is 2 s2 \, \text{s}, time to land on the roof is 3 s3 \, \text{s}, and g=10 m s−2g = 10 \, \text{m s}^{-2}.

Find: The height HH of the building.

From vertical motion, the time to reach maximum height is related to the initial vertical component by

uy=gtu_y = gt

So,

uy=10×2=20 m s−1u_y = 10 \times 2 = 20 \, \text{m s}^{-1}

Now use the vertical displacement after t=3 st = 3 \, \text{s}:

H=uyt−12gt2H = u_y t - \frac{1}{2}gt^2

Substituting the values,

H=(20×3)−12×10×(3)2H = (20 \times 3) - \frac{1}{2} \times 10 \times (3)^2 H=60−45H = 60 - 45 H=15 mH = 15 \, \text{m}

Therefore, the height of the building is 15 m15 \, \text{m}. The correct option is C.

Using vertical displacement in projectile motion

Given: The projectile reaches maximum height in 2 s2 \, \text{s} and reaches the roof in 3 s3 \, \text{s}.

Find: The building height HH.

The hint is that maximum-height time depends only on vertical motion. Therefore first determine the initial vertical velocity.

At the highest point, vertical velocity becomes zero. Hence,

0=uy−gt0 = u_y - gt uy=gtu_y = gt

For t=2 st = 2 \, \text{s},

uy=10×2=20 m s−1u_y = 10 \times 2 = 20 \, \text{m s}^{-1}

Next, after 3 s3 \, \text{s}, the vertical displacement is the roof height:

y=uyt−12gt2y = u_y t - \frac{1}{2}gt^2 H=20×3−12×10×9H = 20 \times 3 - \frac{1}{2} \times 10 \times 9 H=60−45=15 mH = 60 - 45 = 15 \, \text{m}

Therefore, the required height is 15 m15 \, \text{m}.

Common mistakes

  • Using the total initial speed instead of the vertical component. The time to reach maximum height depends only on vertical motion. First find uyu_y from uy=gtu_y = gt, then use it in the displacement equation.

  • Taking the landing time as 2 s2 \, \text{s} instead of 3 s3 \, \text{s}. The value 2 s2 \, \text{s} is only the time to maximum height, whereas the roof is reached after 3 s3 \, \text{s}.

  • Using the wrong sign for gravity in the vertical displacement equation. Since gravity acts downward, the correct expression is y=uyt−12gt2y = u_y t - \frac{1}{2}gt^2, not with a plus sign.

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