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# A ball is thrown forward from the top of a cliff with a velocity of 10ms. The height of the cliff above the ground is 45m. Calculate the distance of the ball to the hirizontal just before hitting the ground

10 months ago

Saurabh Koranglekar
10233 Points
Dear student

Time for fall = sqrt( 2*h/g) = sqrt ( 2*45/10)= 3 sec

Horizontal distance travelled by the ball= 3 * speed horizontal
= 3* 10 = 30 m

Regards
10 months ago
Vikas TU
13786 Points

Dear student
Horizontal velocity is irrelevant. Horizontal and vertical motion are independent of each other.
h = Vin*t + 1/2 gt^2
h = 1/2 gt^2
From here
t = sqrt(2h/g)
t = 3sec
Horizontal distance travelled by the ball= 3 * speed horizontal= 3* 10 = 30 m
10 months ago
aswanth nayak
100 Points

Dear Student,

Horizontal and vertical motion are independent of each other.
h = Vin*t + 1/2 gt^2
h = 1/2 gt^2
From here
t = sqrt(2h/g)
t = 3sec
Horizontal distance travelled by the ball= 3 * speed horizontal= 3* 10 = 30 m
Horizontal velocity is irrelevant.

Regards
8 months ago
Sandeep
103 Points

Time for fall = sqrt( 2*h/g) = sqrt ( 2*45/10)= 3 secHorizontal distance travelled by the ball= 3 * speed horizontal= 3* 10 = 30 m
5 months ago
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### Course Features

• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions